Let $$f(x,y)=\cases{0 & if $x=y$\\7x-3y & otherwise }.$$ Show $f_1(0,0) = 7$ and $f_2(0,0) = -3$, but $f$ is not differentiable at $(0,0)$.
I found $f_1$ and $f_2$ by definition of partial derivative using the limit and found the values. I also reckoned that $f$ is continuous at $(0,0)$. So how can I argue that $f$ is not differentiable? Is that because $f_1$ is not the same as $f_2$? Please clear my confusion.
The fact that the partial derivatives exist doesn't necessarily mean that the function is differentiable. You have to check that there is no such linear map $A : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that:
$$\lim\limits_{||h|| \rightarrow 0} \frac{||f(h) - f(\textbf{0}) - Ah||}{||h||} = 0$$