Find the directional derivative of function $$f(x,y) = 
    \begin{cases}
    \displaystyle
    \frac{x^{3}-3xy^{2}}{x^{2}+y^{2}}, 
    & (x,y) \neq (0,0) \\
    0 , & (x,y) = (0,0)
    \end{cases}$$
Find the directional derivative in $(0,0)$ such that makes an angle $135°$ with $x$ positive direction.
My attempt:
$\lim_{h\to 0}f_x(0+h,0)=\frac{\frac{(0+h)^3-3(0+h)0^2}{(0+h)^2+0^2}-f(0,0)}{h}=1$
$\lim_{h\to 0}f_y(0,0+h)=\frac{\frac{(0)^3-3(0)(0+h)^2}{(0)^2+(0+h)^2}-f(0,0)}{h}=0$
The directional derivative is $(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$, i have to normalize the vector , so $(-1,1)$.
The directional derivative is $(1,0)(-1,1)=-1$
My solution isn't correct , I can't get why , hope for some help , thanks !

Let $v=(v_1,v_2)$ a direction of $\mathbb{R}^2$:
$$\dfrac{\partial f}{\partial v}(0,0)=\lim_{t \to 0}\dfrac{f(0+tv_1,0+tv_2)-f(0,0)}{t}=\lim_{t \to 0}\dfrac{1}{t}\dfrac{(tv_1)^3-3tv_1(tv_2)^2}{(tv_1)^2+(tv_2)^2}=\dfrac{v_1^3-3v_1v_2^2}{v_1^2+v_2^2}$$
Now you can substituite.
Alternatively, you check if $f$ is differentiable in $(0,0)$; if so you can use the following formula:
$$\dfrac{\partial f}{\partial v}(x_0,y_0)=\langle\nabla f(x_0,y_0),v\rangle$$
