[proofplan]
We restrict $f$ and $g$ to the coordinate line through $a$ in the $x_i$ direction. Openness of $U$ ensures that these one-variable restrictions are defined on an interval around $0$. The assumed partial derivatives are exactly the ordinary derivatives of these restrictions at $0$, so the linearity formula follows from the corresponding algebra of difference quotients. For the product rule, we first record that the coordinate-line restriction of $f$ is continuous at $0$, then split the product difference quotient into two terms and pass to the limit.
[/proofplan]
[step:Restrict the functions to the coordinate line through $a$]
Let $e_i\in\mathbb{R}^m$ denote the $i$-th standard basis vector. Since $U$ is open and $a\in U$, there exists $\rho>0$ such that the open ball $B(a,\rho):=\{x\in\mathbb{R}^m:|x-a|<\rho\}$ satisfies $B(a,\rho)\subset U$. If $h\in\mathbb{R}$ satisfies $|h|<\rho$, then $|a+h e_i-a|=|h|<\rho$, so $a+h e_i\in U$.
Define the open interval $I\subset\mathbb{R}$ by $I=(-\rho,\rho)$, and define the one-variable restrictions
\begin{align*}
F:I\to\mathbb{R},\qquad h\mapsto f(a+h e_i).
\end{align*}
\begin{align*}
G:I\to\mathbb{R},\qquad h\mapsto g(a+h e_i).
\end{align*}
By the definition of the [partial derivative](/page/Partial%20Derivative) in the coordinate direction $e_i$, the hypotheses say that the following limits exist:
\begin{align*}
F'(0)=\lim_{h\to 0}\frac{F(h)-F(0)}{h}=\partial_{x_i}f(a).
\end{align*}
\begin{align*}
G'(0)=\lim_{h\to 0}\frac{G(h)-G(0)}{h}=\partial_{x_i}g(a).
\end{align*}
[/step]
[step:Pass the linear combination difference quotient to the limit]
Define $H:I\to\mathbb{R}$ by
\begin{align*}
H(h)=\lambda F(h)+\mu G(h).
\end{align*}
Then $H(h)=(\lambda f+\mu g)(a+h e_i)$ for every $h\in I$. For $h\in I\setminus\{0\}$,
\begin{align*}
\frac{H(h)-H(0)}{h}=\lambda\frac{F(h)-F(0)}{h}+\mu\frac{G(h)-G(0)}{h}.
\end{align*}
Taking the limit as $h\to 0$ and using the two limits established above gives
\begin{align*}
H'(0)=\lambda F'(0)+\mu G'(0)=\lambda\partial_{x_i}f(a)+\mu\partial_{x_i}g(a).
\end{align*}
Since $H$ is the coordinate-line restriction of $\lambda f+\mu g$, this is precisely
\begin{align*}
\partial_{x_i}(\lambda f+\mu g)(a)=\lambda\partial_{x_i}f(a)+\mu\partial_{x_i}g(a).
\end{align*}
[/step]
[step:Derive the product rule from the product difference quotient]
Define $P:I\to\mathbb{R}$ by
\begin{align*}
P(h)=F(h)G(h).
\end{align*}
Then $P(h)=(fg)(a+h e_i)$ for every $h\in I$.
First note that $F$ is continuous at $0$. Indeed, define the difference quotient map $Q_F:I\setminus\{0\}\to\mathbb{R}$ by
\begin{align*}
Q_F(h)=\frac{F(h)-F(0)}{h}.
\end{align*}
Since $Q_F(h)\to F'(0)$ as $h\to 0$, we have
\begin{align*}
F(h)-F(0)=hQ_F(h)\to 0.
\end{align*}
Thus $F(h)\to F(0)$ as $h\to 0$.
For $h\in I\setminus\{0\}$, add and subtract $F(h)G(0)$ in the numerator:
\begin{align*}
\frac{P(h)-P(0)}{h}=F(h)\frac{G(h)-G(0)}{h}+G(0)\frac{F(h)-F(0)}{h}.
\end{align*}
Passing to the limit as $h\to 0$ gives
\begin{align*}
P'(0)=F(0)G'(0)+G(0)F'(0).
\end{align*}
Substituting $F(0)=f(a)$, $G(0)=g(a)$, $F'(0)=\partial_{x_i}f(a)$, and $G'(0)=\partial_{x_i}g(a)$ yields
\begin{align*}
P'(0)=f(a)\partial_{x_i}g(a)+g(a)\partial_{x_i}f(a).
\end{align*}
Since $P$ is the coordinate-line restriction of $fg$, this proves
\begin{align*}
\partial_{x_i}(fg)(a)=f(a)\partial_{x_i}g(a)+g(a)\partial_{x_i}f(a).
\end{align*}
[guided]
Let $e_i\in\mathbb{R}^m$ denote the $i$-th standard basis vector. Choose $\rho>0$ such that $B(a,\rho)\subset U$, and define the open interval $I\subset\mathbb{R}$ by $I=(-\rho,\rho)$. Define the coordinate-line restrictions $F:I\to\mathbb{R}$ and $G:I\to\mathbb{R}$ by
\begin{align*}
F(h)=f(a+h e_i).
\end{align*}
\begin{align*}
G(h)=g(a+h e_i).
\end{align*}
By the definition of the partial derivative in the coordinate direction $e_i$, the assumed existence of $\partial_{x_i}f(a)$ and $\partial_{x_i}g(a)$ means that $F'(0)$ and $G'(0)$ exist and satisfy
\begin{align*}
F'(0)=\partial_{x_i}f(a).
\end{align*}
\begin{align*}
G'(0)=\partial_{x_i}g(a).
\end{align*}
The product rule has one extra point that the linearity argument does not need: in the term where the difference quotient of $G$ appears, the coefficient is $F(h)$, not $F(0)$. We therefore first prove that $F(h)\to F(0)$ as $h\to 0$ using only the assumed existence of $F'(0)$.
For $h\in I\setminus\{0\}$, define the difference quotient map $Q_F:I\setminus\{0\}\to\mathbb{R}$ by
\begin{align*}
Q_F(h)=\frac{F(h)-F(0)}{h}.
\end{align*}
The hypothesis that $\partial_{x_i}f(a)$ exists means exactly that $Q_F(h)$ has a finite limit as $h\to 0$, namely $F'(0)=\partial_{x_i}f(a)$. Hence $Q_F$ is bounded near $0$: there are constants $\delta>0$ and $M>0$ such that $|Q_F(h)|\le M$ whenever $0<|h|<\delta$. For such $h$,
\begin{align*}
|F(h)-F(0)|=|h|\,|Q_F(h)|\le M|h|.
\end{align*}
Since $M|h|\to 0$ as $h\to 0$, the squeeze property gives $F(h)\to F(0)$.
Now define $P:I\to\mathbb{R}$ by
\begin{align*}
P(h)=F(h)G(h).
\end{align*}
This is the coordinate-line restriction of the product $fg$, because $P(h)=f(a+h e_i)g(a+h e_i)=(fg)(a+h e_i)$. For $h\in I\setminus\{0\}$, we split the product difference by adding and subtracting the intermediate term $F(h)G(0)$:
\begin{align*}
P(h)-P(0)=F(h)G(h)-F(0)G(0).
\end{align*}
\begin{align*}
P(h)-P(0)=F(h)\bigl(G(h)-G(0)\bigr)+G(0)\bigl(F(h)-F(0)\bigr).
\end{align*}
Dividing by $h$ gives
\begin{align*}
\frac{P(h)-P(0)}{h}=F(h)\frac{G(h)-G(0)}{h}+G(0)\frac{F(h)-F(0)}{h}.
\end{align*}
As $h\to 0$, the first factor in the first term satisfies $F(h)\to F(0)$, while the second factor satisfies
\begin{align*}
\frac{G(h)-G(0)}{h}\to G'(0).
\end{align*}
The second term satisfies
\begin{align*}
G(0)\frac{F(h)-F(0)}{h}\to G(0)F'(0).
\end{align*}
Therefore
\begin{align*}
P'(0)=F(0)G'(0)+G(0)F'(0).
\end{align*}
Finally, translating back from the coordinate-line notation gives $F(0)=f(a)$, $G(0)=g(a)$, $F'(0)=\partial_{x_i}f(a)$, and $G'(0)=\partial_{x_i}g(a)$. Thus
\begin{align*}
\partial_{x_i}(fg)(a)=f(a)\partial_{x_i}g(a)+g(a)\partial_{x_i}f(a).
\end{align*}
[/guided]
[/step]
