= Solution
Differentiability means existence of a <linear map> $A:\mathbb R^2\to\mathbb R$ such that
$$
f(x_0+h,y_0+k)=f(x_0,y_0)+A(h,k)+r(h,k),\qquad\frac{r(h,k)}{\sqrt{h^2+k^2}}\longrightarrow0.
$$
This is the <Fréchet derivative>. Substituting increments on the coordinate axes shows $A(h,k)=f_x(x_0,y_0)h+f_y(x_0,y_0)k$. For the unit direction $v=(\cos\alpha,\sin\alpha)$, the remainder is $r(tv)=o(|t|)$. Dividing the increment by $t$ proves differentiability of the restricted function and gives its <directional derivative>:
$$
\boxed{g_\alpha'(0)=f_x(x_0,y_0)\cos\alpha+f_y(x_0,y_0)\sin\alpha.}
$$
For the supplied rational function, both coordinate-axis restrictions are zero, so $f_x(0,0)=f_y(0,0)=0$. Any <Fréchet derivative> would therefore be the zero map. But $f(t,t)=t$, and hence $|f(t,t)|/\sqrt{t^2+t^2}=1/\sqrt2$ for nonzero $t$. This contradicts the required vanishing remainder. Thus
$$
\boxed{f\text{ is not differentiable at }(0,0).}
$$
In fact every <directional derivative> exists: along $v$ it equals $\cos^2\alpha\sin\alpha+\cos\alpha\sin^2\alpha$. These values do not depend linearly on the direction, illustrating why existence of directional derivatives is weaker than differentiability.
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