Define $ f:\mathbb R^2\to\mathbb R $ by
$ f(x,y) := \begin{cases} \frac{2 x^2 -4 x + 6 y^3 - 13 y^2 + 8 y +1}{x- y}, \quad &\text{if }x\neq y; \\ 0,\quad &\text{if }x=y. \end{cases} $
I'm trying to solve for $\frac{\partial f}{\partial x}(1,1) $ and $\frac{\partial f}{\partial y}(1,1) $.
I've tried to solve it a few times but I'm not quite sure if I'm solving it correctly. I tried a couple different techniques but I'm not quite sure which one applies and if I'm applying them correctly. I know neither one of them are $0$ but I seem to get that answer quite often for the partial derivative with respect to $x$.
Technique 1:
$$\lim_{x\to1}\frac{f(x,1)-f(1,1)}{x-1}=\lim_{x\to1}\frac{\frac{2x^2-4x+2}{x-1}}{x-1}=\lim_{x\to1}\frac{\frac{2(x-1)^2}{x-1}}{x-1}=\lim_{x\to1}\frac{2(x-1)}{x-1}=2$$
$$\lim_{y\to1}\frac{f(1,y)-f(1,1)}{y-1}=\lim_{y\to1}\frac{\frac{6y^3-13y^2+8y-1}{1-y}}{y-1}=\lim_{y\to1}\frac{\frac{(y-1)^2(6y-1)}{1-y}}{y-1}=\lim_{y\to1}\frac{(y-1)^2(6y-1)}{(1-y)(y-1)}=\lim_{y\to1}\frac{(y-1)(6y-1)}{1-y}=\lim_{y\to1}-6y+1=-5$$
The other technique is that I first take the partial derivative to $x$ from the function and call it $g(x)$ for clarity. Then I set $y=1$ in $g(x)$ (the partial derivative). Then I solve the limit of $x$ approaching $1$ for $g(x)$.
I would really appreciate any help on fixing my solution for this. If it's correct, can someone explain why exactly it's correct and how I can check this. Also are both techniques equivalent?
Thanks $:)$