Partial derivatives: Why do results vary?

Question

When I attempt to compute $f_{y}(0,0)$, I first set $x = 0$ such that $f(0,y) = \frac{y^2}{y^2} = 1$, and so $f_{y}(0,y) = 0$. So its passes differentiability w.r.t.y near $(0,0,f(0,0))$.

However, if I compute this exact partial derivative using the definition of differentiation:

$\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}$, I end up obtaining $\lim_{h \to 0}\frac{1}{h}$ which reveals that $f_{y}(0,0)$ does not exist.

How can such contradicting result be explained?

Answer 1

The function $$ g(y)= f(0,y) = \begin{cases} 1, & y\neq 0,\\ 0, & y=0 \end{cases} $$ isn't continuous at $y=0$, hence not differentiable there either, so $g'(0)=f_y(0,0)$ doesn't exist.

Answer 2

In this case you really need to use the definition, not the usual rules for derivatives. This comes from the fact that the correct expression for $f(0,y)$ is

$$ f(0,y)=\begin{cases} 1, &y \ne 0 \\ 0, &y=0 \end{cases} $$

Answer 3

$ \frac{f(0,h)-f(0,0)}{h}= \frac{1}{h}.$ This shows that $ \lim_{h \to 0}\frac{f(0,h)-f(0,0)}{h}$ does not exist. Hence $f_y(0,0)$ does not exist.