Given a function defined by : $$f(x,y) = \begin{cases} \frac{x^{3}+y^{3}}{x^{2}+y^{2}} & (x,y) \ne (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$
Find $f_x(0,0),f_y(0,0)$ is they exist ,is the function continuous at $(0,0)$?
$f_x(0,0)$ and $f_y(0,0)$ exist if the following limits exist (respectively):
$$\lim_{\Delta x \to 0}\frac{f\left(\Delta x,0\right)-f\left(0,0\right)}{ \Delta x}=\lim_{\Delta x \to 0}\frac{\Delta x-0}{\Delta x}=1$$ $$\lim_{\Delta y \to 0}\frac{f\left(0,\Delta y\right)-f\left(0,0\right)}{ \Delta y}=\lim_{\Delta y \to 0}\frac{\Delta y-0}{\Delta y}=1$$
So $f_x(0,0) =1=f_y(0,0)$
About the continuity of the function at the given point we have that $$\lim_{(x,y) \to (0,0)} \frac{x^{3}+y^{3}}{x^{2}+y^{2}}=0$$
Because for $\left|x\right|,\left|y\right|<\delta$ :
$$\left|\frac{x^{3}+y^{3}}{x^{2}+y^{2}}-0\right|\le\frac{\left|x\right|^{3}+\left|y\right|^{3}}{x^{2}+y^{2}}=\frac{\left(\left|x\right|+\left|y\right|\right)\left(x^{2}-\left|xy\right|+y^{2}\right)}{x^{2}+y^{2}}\le\frac{\left(\left|x\right|+\left|y\right|\right)\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$$$$=\left|x\right|+\left|y\right|<2\delta$$
So taking $\delta \le \frac{\epsilon}{2}$ proves the claim.
I would like to know if my solution is true or not,and see the errors (if exists),so if you have an alternative solution then please first check mine and then provides yours.
Your solution is fine. For the continuity of $f$ in $(0,0)$ you can use polar coordinates:
With $x= r \cos t$ and $y=r \sin t$ you get
$$|f(x,y)-0| \le r (| \cos t|+ |\sin t|) \le 2r=2 \sqrt{x^2+y^2}.$$