Let be $f:\mathbb{R}^2\to\mathbb{R}$, where$$f(x,y):=\begin{cases}x,&x\neq y\\1,&x=y, x\neq 0,y\neq 0\\0,&x=y=0\end{cases}$$
Show that $f$ is not differentiable but partially differentiable at $(0,0)$.
My approach:
We immediately see that\begin{align*}&\text{if }x\neq 0\implies \frac{f(x,0)-f(0,0)}{x-0}=\frac{x-0}{x}=1\implies \lim\limits_{x\to0;x\neq 0}\frac{f(x,0)-f(0,0)}{x-0}=1\\&\text{if }y\neq 0\implies\frac{f(0,y)-f(0,0)}{y-0}=\frac{0-0}{y}=0\implies \lim\limits_{y\to0;y\neq0}\frac{f(0,y)-f(0,0)}{y-0}=0.\end{align*}
So both partials exists at point $(0,0)$. However,\begin{align*}\frac{\left|f(t,t)-f(0,0)-A\left({t\choose t}-{0\choose 0}\right)\right|}{\Vert {t\choose t}-{0\choose 0}\Vert_\infty}=\frac{|1-0-a_1t-a_2t|}{|t|}\geq\left|\left|\frac{1}{t}\right|-|a_1+a_2|\right|,\end{align*}where $A={a_1\choose a_2}$ is an arbitrary but fixed matrix. Letting $t\to0$ shows that the limit is infinite, so that $f$ can't be differentiable at $(0,0)$.
Is this correct?