[proofplan] Fix an admissible two-sided variation direction $h$ at $y$ and reduce the functional problem to a one-variable problem along the affine variation curve $\varepsilon\mapsto y+\varepsilon h$. The local extremum assumption on $J$ implies that the induced real-valued function has a local extremum at $\varepsilon=0$. The ordinary one-variable necessary condition for differentiable local extrema gives vanishing derivative at $0$, and the definition of first variation identifies that derivative with $\delta J[y;h]$. [/proofplan] [step:Restrict the functional to the admissible variation line] Let $h\in\mathcal V$ be an admissible two-sided variation direction at $y$. By admissibility, there exists $\rho_h>0$ such that $y+\varepsilon h\in\mathcal A$ for every $\varepsilon\in(-\rho_h,\rho_h)$. Define the one-variable variation map \begin{align*} \phi:(-\rho_h,\rho_h)&\to\mathbb R \end{align*} by \begin{align*} \phi(\varepsilon)=J[y+\varepsilon h]. \end{align*} By the assumed existence of the first variation in the direction $h$, the derivative $\phi'(0)$ exists and satisfies \begin{align*} \phi'(0)=\delta J[y;h]. \end{align*} [guided] We fix one admissible two-sided variation direction $h\in\mathcal V$ at $y$. The phrase “two-sided” is the point that makes the ordinary derivative test applicable: it gives a whole interval around $0$, not just perturbations with $\varepsilon\geq0$ or $\varepsilon\leq0$. Thus there exists $\rho_h>0$ such that $y+\varepsilon h\in\mathcal A$ for every $\varepsilon\in(-\rho_h,\rho_h)$. Now define the associated real-valued one-variable function \begin{align*} \phi:(-\rho_h,\rho_h)&\to\mathbb R \end{align*} by \begin{align*} \phi(\varepsilon)=J[y+\varepsilon h]. \end{align*} This function records the value of the functional $J$ along the admissible line through $y$ in the direction $h$. Since the first variation of $J$ at $y$ in the direction $h$ exists by hypothesis, the derivative of this one-variable function at $0$ exists, and the definition of first variation gives \begin{align*} \phi'(0)=\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}J[y+\varepsilon h]=\delta J[y;h]. \end{align*} [/guided] [/step] [step:Transfer the local extremum from $J$ to the one-variable function] Assume first that $y$ is a local minimizer relative to admissible two-sided variations. Then, after decreasing $\rho_h$ if necessary, we have \begin{align*} J[y]\leq J[y+\varepsilon h] \end{align*} for all $\varepsilon\in(-\rho_h,\rho_h)$. Hence \begin{align*} \phi(0)\leq \phi(\varepsilon) \end{align*} for all $\varepsilon\in(-\rho_h,\rho_h)$, so $0$ is a local minimizer of $\phi$. If $y$ is instead a local maximizer, the same argument gives \begin{align*} \phi(0)\geq \phi(\varepsilon) \end{align*} for all sufficiently small $\varepsilon$, so $0$ is a local maximizer of $\phi$. [/step] [step:Apply the one-variable derivative test at the interior extremum] In either case, $0$ is an interior point of the interval $(-\rho_h,\rho_h)$ and $\phi$ is differentiable at $0$. By Fermat's one-variable necessary condition for differentiable local extrema, $\phi'(0)=0$. Combining this with the identity obtained from the definition of first variation gives \begin{align*} \delta J[y;h]=\phi'(0)=0. \end{align*} Because $h$ was an arbitrary admissible two-sided variation direction at $y$, the conclusion holds for every such $h$. [/step]