[proofplan]
We first prove the corresponding untwisted statement on $T^*(X\times Y)$ for the map sending $(x,y,\theta)$ to the covector $d_{x,y}\phi$. The nondegeneracy hypothesis makes the critical set $\Sigma_\phi$ a smooth conic manifold of the correct dimension, and the standard nondegenerate phase rank argument shows that the phase map is an immersion. Pulling back the canonical one-form gives $d(\phi|_{\Sigma_\phi})$, because the $\theta$ derivatives vanish on $\Sigma_\phi$; hence the pulled-back symplectic form is zero. Since the immersed image is isotropic of half dimension, it is Lagrangian, and the sign change in the $Y$ covector converts the canonical product symplectic form into $\pi_X^*\omega_X-\pi_Y^*\omega_Y$.
[/proofplan]
[step:Pass to the untwisted cotangent bundle of $X\times Y$]
Let $M := X\times Y$, and write a point of $M$ as $z=(x,y)$. In a local coordinate chart on $M$, write $z=(z_1,\dots,z_m)$, where $m=\dim X+\dim Y$. Define the untwisted phase map
\begin{align*}
I_\phi:\Sigma_\phi &\to T^*M
\end{align*}
\begin{align*}
(z,\theta) &\mapsto (z,\partial_z\phi(z,\theta)).
\end{align*}
Here $\partial_z\phi(z,\theta)\in T_z^*M$ is the covector whose local coordinate expression is
\begin{align*}
\partial_z\phi(z,\theta)=\sum_{a=1}^{m}\partial_{z_a}\phi(z,\theta)\,dz_a.
\end{align*}
The assumptions $\partial_x\phi\neq 0$ and $\partial_y\phi\neq 0$ imply that $I_\phi(\Sigma_\phi)\subset T^*M\setminus 0$, and also that the twisted map $F_\phi$ lands in $(T^*X\setminus 0)\times(T^*Y\setminus 0)$.
[/step]
[step:Use the regular level set theorem to make $\Sigma_\phi$ a smooth conic manifold]
Define the constraint map
\begin{align*}
G:\Omega &\to \mathbb{R}^N
\end{align*}
\begin{align*}
(z,\theta) &\mapsto \bigl(\partial_{\theta_1}\phi(z,\theta),\dots,\partial_{\theta_N}\phi(z,\theta)\bigr).
\end{align*}
Then $\Sigma_\phi=G^{-1}(0)$. At every point of $\Sigma_\phi$, the differential
\begin{align*}
dG_{(z,\theta)}:T_{(z,\theta)}\Omega \to \mathbb{R}^N
\end{align*}
has rank $N$, because its component covectors are precisely
\begin{align*}
d(\partial_{\theta_1}\phi),\dots,d(\partial_{\theta_N}\phi).
\end{align*}
Thus $0$ is a regular value of $G$ along $G^{-1}(0)$. By the regular level set theorem, $\Sigma_\phi$ is a smooth submanifold of $\Omega$ of codimension $N$. Therefore
\begin{align*}
\dim \Sigma_\phi=\dim X+\dim Y.
\end{align*}
Since $\Omega$ is conic in $\theta$ and $\phi$ is homogeneous of degree $1$ in $\theta$, each $\partial_{\theta_j}\phi$ is homogeneous of degree $0$ in $\theta$. Hence $G(z,\lambda\theta)=G(z,\theta)$ for every $\lambda>0$ for which $(z,\lambda\theta)\in\Omega$, so $\Sigma_\phi$ is conic.
[/step]
[step:Prove that the untwisted phase map is an immersion]
Let $(z,\theta)\in\Sigma_\phi$, and let $v\in T_{(z,\theta)}\Sigma_\phi$ satisfy
\begin{align*}
d(I_\phi)_{(z,\theta)}(v)=0.
\end{align*}
Since the base component of $I_\phi$ is the projection $(z,\theta)\mapsto z$, this implies
\begin{align*}
dz_a(v)=0
\end{align*}
for every $1\leq a\leq m$. Since the fiber component of $I_\phi$ is $\partial_z\phi$, it also implies
\begin{align*}
d(\partial_{z_a}\phi)_{(z,\theta)}(v)=0
\end{align*}
for every $1\leq a\leq m$. Finally, because $v\in T_{(z,\theta)}\Sigma_\phi$ and $\Sigma_\phi=G^{-1}(0)$, we have
\begin{align*}
d(\partial_{\theta_j}\phi)_{(z,\theta)}(v)=0
\end{align*}
for every $1\leq j\leq N$.
The identities $dz_a(v)=0$ show that $v$ is vertical in the $\theta$ direction. Write
\begin{align*}
v=\sum_{k=1}^{N}v_k\,\partial_{\theta_k}
\end{align*}
in the chosen local coordinates. The equalities
\begin{align*}
d(\partial_{z_a}\phi)(v)=0
\end{align*}
and
\begin{align*}
d(\partial_{\theta_j}\phi)(v)=0
\end{align*}
together say that the differential of the full gradient $d\phi$ in the vertical direction $v$ is zero. Equivalently,
\begin{align*}
\sum_{k=1}^{N}v_k\,\partial_{\theta_k}\partial_{z_a}\phi=0
\end{align*}
for every $a$, and
\begin{align*}
\sum_{k=1}^{N}v_k\,\partial_{\theta_k}\partial_{\theta_j}\phi=0
\end{align*}
for every $j$.
Since $\phi$ is homogeneous of degree $1$ in $\theta$, Euler's homogeneous identity gives
\begin{align*}
\phi(z,\theta)=\sum_{k=1}^{N}\theta_k\,\partial_{\theta_k}\phi(z,\theta).
\end{align*}
Differentiating this identity in the vertical direction $v$ and using $\partial_{\theta_k}\phi=0$ on $\Sigma_\phi$ gives no further condition on $\phi$ itself, but differentiating the identities for all first derivatives shows that the covector
\begin{align*}
\sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)_{(z,\theta)}
\end{align*}
vanishes on every coordinate vector $\partial_{z_a}$ and on every coordinate vector $\partial_{\theta_j}$. Hence
\begin{align*}
\sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)_{(z,\theta)}=0
\end{align*}
as an element of $T^*_{(z,\theta)}\Omega$. By the assumed [linear independence](/page/Linear%20Independence) of these covectors, $v_k=0$ for every $k$. Thus $v=0$, so $d(I_\phi)_{(z,\theta)}$ is injective. Therefore $I_\phi$ is an immersion.
[guided]
The only delicate point is to show that the rank condition on the phase really forces the phase map to be immersive. Fix $(z,\theta)\in\Sigma_\phi$ and suppose that a tangent vector $v\in T_{(z,\theta)}\Sigma_\phi$ is killed by $d(I_\phi)_{(z,\theta)}$. The map $I_\phi$ has two components: the base point $z$ and the covector $\partial_z\phi$. Therefore $d(I_\phi)(v)=0$ gives two families of equations:
\begin{align*}
dz_a(v)=0
\end{align*}
for every base coordinate $z_a$, and
\begin{align*}
d(\partial_{z_a}\phi)(v)=0
\end{align*}
for every $1\leq a\leq m$.
The first family means that $v$ has no $z$ component. Thus, in local coordinates, it has the form
\begin{align*}
v=\sum_{k=1}^{N}v_k\,\partial_{\theta_k}.
\end{align*}
Since $v$ is tangent to $\Sigma_\phi=G^{-1}(0)$, it also satisfies the linearized constraint equations
\begin{align*}
d(\partial_{\theta_j}\phi)(v)=0
\end{align*}
for every $1\leq j\leq N$.
Now combine the two families. The equations $d(\partial_{z_a}\phi)(v)=0$ say that the covector
\begin{align*}
\sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)
\end{align*}
vanishes on the coordinate vector $\partial_{z_a}$, because mixed partial derivatives agree:
\begin{align*}
\sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)(\partial_{z_a})
=
\sum_{k=1}^{N}v_k\,\partial_{z_a}\partial_{\theta_k}\phi
=
d(\partial_{z_a}\phi)(v)
=
0.
\end{align*}
The tangent-to-$\Sigma_\phi$ equations say that the same covector vanishes on each coordinate vector $\partial_{\theta_j}$:
\begin{align*}
\sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)(\partial_{\theta_j})
=
d(\partial_{\theta_j}\phi)(v)
=
0.
\end{align*}
Since the coordinate vectors $\partial_{z_a}$ and $\partial_{\theta_j}$ span $T_{(z,\theta)}\Omega$, we obtain
\begin{align*}
\sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)_{(z,\theta)}=0.
\end{align*}
The nondegeneracy hypothesis is exactly the statement that the covectors $d(\partial_{\theta_1}\phi),\dots,d(\partial_{\theta_N}\phi)$ are linearly independent on $\Sigma_\phi$. Therefore every coefficient $v_k$ is zero. Hence $v=0$, proving that $d(I_\phi)$ is injective at $(z,\theta)$. Since the point was arbitrary, $I_\phi$ is an immersion.
[/guided]
[/step]
[step:Pull back the canonical one-form and prove isotropicity]
Let $\alpha_M$ denote the canonical one-form on $T^*M$, with local expression
\begin{align*}
\alpha_M=\sum_{a=1}^{m}\xi_a\,dz_a
\end{align*}
in cotangent coordinates $(z,\xi)$. We use the convention
\begin{align*}
\omega_M=d\alpha_M
\end{align*}
for the canonical symplectic form.
Pulling $\alpha_M$ back along $I_\phi$ gives
\begin{align*}
I_\phi^*\alpha_M=\sum_{a=1}^{m}\partial_{z_a}\phi\,dz_a.
\end{align*}
On $\Sigma_\phi$, the defining equations give $\partial_{\theta_j}\phi=0$ for every $j$. Hence the differential of the restricted function
\begin{align*}
\phi|_{\Sigma_\phi}:\Sigma_\phi\to\mathbb{R}
\end{align*}
is
\begin{align*}
d(\phi|_{\Sigma_\phi})
=
\sum_{a=1}^{m}\partial_{z_a}\phi\,dz_a
+
\sum_{j=1}^{N}\partial_{\theta_j}\phi\,d\theta_j
=
\sum_{a=1}^{m}\partial_{z_a}\phi\,dz_a.
\end{align*}
Therefore
\begin{align*}
I_\phi^*\alpha_M=d(\phi|_{\Sigma_\phi}).
\end{align*}
Taking exterior derivatives and using $d^2=0$, we obtain
\begin{align*}
I_\phi^*\omega_M
=
I_\phi^*(d\alpha_M)
=
d(I_\phi^*\alpha_M)
=
d^2(\phi|_{\Sigma_\phi})
=
0.
\end{align*}
Thus $I_\phi(\Sigma_\phi)$ is isotropic with its immersed structure.
[/step]
[step:Use the dimension count to conclude the untwisted image is Lagrangian]
The symplectic manifold $T^*M$ has dimension
\begin{align*}
2\dim M=2(\dim X+\dim Y).
\end{align*}
The domain $\Sigma_\phi$ has dimension
\begin{align*}
\dim \Sigma_\phi=\dim X+\dim Y=\frac{1}{2}\dim T^*M.
\end{align*}
Since $I_\phi$ is an immersion and $I_\phi^*\omega_M=0$, the immersed tangent space
\begin{align*}
d(I_\phi)_{(z,\theta)}(T_{(z,\theta)}\Sigma_\phi)
\end{align*}
is an isotropic subspace of $T_{I_\phi(z,\theta)}T^*M$ of maximal possible dimension. Hence it is a Lagrangian subspace. Therefore $I_\phi:\Sigma_\phi\to T^*M$ is a conic Lagrangian immersion, and its immersed image is a conic immersed Lagrangian submanifold of $T^*M\setminus 0$.
[/step]
[step:Apply the cotangent sign change on the $Y$ factor]
Use the product identification
\begin{align*}
T^*(X\times Y)\cong T^*X\times T^*Y
\end{align*}
given by
\begin{align*}
(x,y;\xi,\eta)\mapsto (x,\xi;y,\eta).
\end{align*}
Define the fiber sign-change map
\begin{align*}
S:T^*X\times T^*Y &\to T^*X\times T^*Y
\end{align*}
\begin{align*}
(x,\xi;y,\eta) &\mapsto (x,\xi;y,-\eta).
\end{align*}
Then
\begin{align*}
F_\phi=S\circ I_\phi.
\end{align*}
Let $\alpha_X$ and $\alpha_Y$ denote the canonical one-forms on $T^*X$ and $T^*Y$, and let
\begin{align*}
\omega_X=d\alpha_X,\qquad \omega_Y=d\alpha_Y.
\end{align*}
Under the product identification, the canonical one-form on $T^*(X\times Y)$ is
\begin{align*}
\alpha_M=\pi_X^*\alpha_X+\pi_Y^*\alpha_Y.
\end{align*}
The map $S$ fixes the $T^*X$ covector and multiplies the $T^*Y$ covector by $-1$, so
\begin{align*}
S^*(\pi_X^*\alpha_X+\pi_Y^*\alpha_Y)
=
\pi_X^*\alpha_X-\pi_Y^*\alpha_Y.
\end{align*}
Taking exterior derivatives gives
\begin{align*}
S^*\omega_M
=
\pi_X^*\omega_X-\pi_Y^*\omega_Y.
\end{align*}
Since $I_\phi$ is Lagrangian for $\omega_M$, the composition $F_\phi=S\circ I_\phi$ is Lagrangian for $\pi_X^*\omega_X-\pi_Y^*\omega_Y$. The conic property follows from the homogeneity of $\phi$: for $\lambda>0$,
\begin{align*}
\partial_x\phi(x,y,\lambda\theta)=\lambda\,\partial_x\phi(x,y,\theta)
\end{align*}
and
\begin{align*}
\partial_y\phi(x,y,\lambda\theta)=\lambda\,\partial_y\phi(x,y,\theta).
\end{align*}
Thus $F_\phi(x,y,\lambda\theta)$ is obtained from $F_\phi(x,y,\theta)$ by positive fiber dilation in both cotangent factors. Therefore $F_\phi$ is a conic Lagrangian immersion, and $C_\phi=F_\phi(\Sigma_\phi)$ is a conic immersed Lagrangian submanifold with the induced immersed structure.
[/step]
