Read "$x=0$ or $y=0 $" as $x=0\cup y=0$
I don't understand the section of the solution highlighted in green.
For instance if: $f(x)=e^{-1/x^2}$ for $x\neq0$
and $f(0)=0 $
I cannot just say $f(0)=0 \implies f'(0)=0 \implies f$ is differentiable at $0$
It has to be shown via the definition of the derivative. So why in this case does the author take such a simplistic approach?
On
The meaning of $x=0$ or $y=0$ is not at one point $(0,0)$. $x=0$ means that you are restricted to the $y$ axis. All along that axis $f(x,y)=0$. Because they have defined $f(x,y)$ to be zero at $(0,0)$ means that the partial derivative there, along the $y$ axis, will be zero. If $x$ is not zero then the denominator is not zero and so the derivative exists as the quotient of two polynomials.
The meaning is that for $x=0$ and every $y$, you have $f(x,y)=0$, i.e. for $x=0$ the induced function on $y$ is constant, thus differentiable. The same goes for $y=0$.