Computing this limit: $ \lim_{y\to0} \frac{f(x,y) - f(x-y,y)}{y} = g(x)$ If $f(x,y) \in \mathbb{R^2}$ and $g(x) \in \mathbb{R}$. Assuming
$\frac{f(x,y) - f(x-y,y)}{y} = g(x); \forall y \in \mathbb{R}$
$$$$
Can we do the following:
$$ \lim_{y\to0} \frac{f(x,y) - f(x-y,y)}{y} = g(x)$$
$$ \lim_{y\to0} \frac{\partial{f(x,y)}}{{\partial x}}= g(x) $$
 A: No, that's not quite valid, though it's getting you to the right answer. The difference quotient inside the limit is not equal to the partial $\frac{\partial f}{\partial x}$, because it's just a difference quotient; it doesn't become a derivative until you take a limit. And even after you do, that quotient is not simply a partial in the $x$ direction, because both terms are changing.
$\begin{align*}
\lim_{y \to 0} \frac{f(x,y) - f(x-y,y)}{y} &= \lim_{y \to 0} \frac{f(x,y) - f(x,0)}{y} - \lim_{y \to 0} \frac{f(x-y,y) - f(x,0)}{y} \\
&= \frac{\partial f}{\partial y}(x,0) - \left(-\frac{\partial f}{\partial x}(x,0) + \frac{\partial f}{\partial y}(x,0)\right) \\
&= \frac{\partial f}{\partial x}(x,0)
\end{align*}$
For the second equality above I recognized those difference quotients as directional derivatives.
A: Why not to use Taylor expansion around $y=0$ ?
$$f(x,y)=f(x,0)+y f^{(0,1)}(x,0)+O\left(y^2\right)$$
$$f(x-y,y)=f(x,0)+y \left(f^{(0,1)}(x,0)-f^{(1,0)}(x,0)\right)+O\left(y^2\right)$$
$$f(x,y)-f(x-y,y)=y f^{(1,0)}(x,0)+O\left(y^2\right)$$
$$\frac{f(x,y)-f(x-y,y) }y=f^{(1,0)}(x,0)+O\left(y^1\right)$$
