Proof of Derivative of a Coordinate Slice of a $C^1$ Function on a Euclidean Open Set

Define

$$\phi:[a,b]\to U,\qquad \phi(t)=(x_1,\dots,x_{i-1},t,x_{i+1},\dots,x_n).$$

This is well defined by the hypothesis that the whole coordinate slice over $[a,b]$ lies in $U$. Then $g=f\circ\phi$ on $[a,b]$. Since $f$ is a $C^1$ map, it is continuous on $U$ and differentiable at each point of $U$ by C^1 Maps on Euclidean Open Sets are Differentiable. Because $\phi$ is continuous, the composition $g=f\circ\phi$ is continuous on $[a,b]$.

Now let $t\in(a,b)$. The map $\phi$ is differentiable at $t$, and its derivative sends $1\in\mathbb{R}$ to the standard basis vector $e_i\in\mathbb{R}^n$. Hence by the chain rule Chain Rule for C^1 Maps Between Euclidean Spaces, the derivative of $g=f\circ\phi$ at $t$ is

$$g'(t)=Df(\phi(t))(e_i).$$

By the definition of the partial derivative Partial Derivative of a Coordinate Function, this is exactly

$$\frac{\partial f}{\partial x_i}(x_1,\dots,x_{i-1},t,x_{i+1},\dots,x_n).$$

This proves the theorem.