 Solution

Differentiability means existence of a linear map $A:{\mathbb{R}}^2\to \mathbb{R}$ such thatThis 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: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 $|f(t,t)|/\sqrt{t^2+t^2}=1/\sqrt{2}$ for nonzero $t$. This contradicts the required vanishing remainder. ThusIn 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.