Let's suppose we have an infinitely differentiable multivariable function, $f(x,y)$. Which
$$f(x,a) = g(x)$$
Does the following hold? Does the order of steps matter: The left: I take the derivative and then evaluate at $y=0$; on the right I evaluate at $y=0$ and then take the derivative.
$$\frac{\partial^n}{\partial{x^n}}f(x,a) = \frac{\partial^n}{\partial{x^n}}g(x)$$
I tried out some and, by the results, I'm attempted to conclude it holds. See below:
$$f(x,y) = xe^y, xe^{y-1}, x^2+y, \ln(x^2-yx), x^y $$
EDIT
So, basically, I tried to prove it by myself, here it is:
First: We only need to prove for the first, because if the first partial are equivalente, then the others will be to.
Second: Let $f = f(x_0,x_1, \cdots, x_i, \cdots, x_n)$ and $g = f(x_0,x_1, \cdots, a, \cdots, x_n)$, then
$$ \frac{\partial{f}}{\partial{x_r}}= \lim_{h \to 0} \frac{f(x_0+h,\cdots,x_i,\cdots,x_n) - f(x_0,\cdots,x_i,\cdots,x_n)}{h}|_{x_i=a} \,\,(1)$$ $$\frac{\partial{f}}{\partial{x_r}}= \lim_{h \to 0} \frac{f(x_0+h,\cdots,a,\cdots,x_n) - f(x_0,\cdots,a,\cdots,x_n)}{h}$$
.
$$ \frac{\partial{g}}{\partial{x_r}}= \lim_{h \to 0} \frac{f(x_0+h,\cdots,a,\cdots,x_n) - f(x_0,\cdots,a,\cdots,x_n)}{h}\,\, (2)$$
Such that $r \neq i$. Then, we substitute $x_i=0$ in (1). Therefore
$$ \frac{\partial{f}}{\partial{x_r}}= \frac{\partial{g}}{\partial{x_r}} $$
$\frac{d}{dx}g(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}=\lim_{h\to0}\frac{f(x+h,0)-f(x,0)}{h}=\frac{\partial}{\partial x}f(x,0)$.
Perform induction on $n$ for the $n^\text{th}$ derivative.