$f$ is continuous at $(0,0)$ as you can prove the inequality $$ \vert f(x,y) \vert \le \frac{\vert xy \vert}{\sqrt{x^2+y^2}} \le \frac{\sqrt{x^2+y^2}}{2}$$
$f$ is not differentiable at $(0,0)$. If it would be the case its derivative would be equal to the zero matrix as the partial derivatives are equal to $0$. Hence all the directional derivatives would vanish which is not the case. A contradiction proving that $f$ is not differentiable at the origin.
Before showing differentiability you need to show continuity in $(0,0)$.
We are in $\mathbb R^2$, so we have to select a norm, and given the denominator of $f(x,y)$ it seems appropriate to choose the Euclidean norm$$||(x,y)||_2=\sqrt{x^2+y^2}$$
So notice $|x^3|<x^2$ and $|y^3|<y^2$ when $x,y$ are small.
$$|f(x,y)|\le\dfrac{|x^3|+|y^3|}{\sqrt{x^2+y^2}}\le \dfrac{x^2+y^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}=||(x,y)||_2\to 0$$
Therefore $f$ is continuous in $(0,0)$ and $f(0,0)=0$.
Now for differentiability you need to evaluate
$\dfrac {|f(x,0)-f(0,0)|}{|x|}=\dfrac{\frac {x^3}{|x|}-0}{|x|}=\dfrac{x^3}{x^2}=|x|\to 0$
By symmetry we have the same for $y$ and set $f_x(0,0)=0$ and $f_y(0,0)=0$.
Now we look at the differentiability:
$\begin{align}\dfrac {|f(0+x,0+y)-\overbrace{f(0,0)}^0-\overbrace{f_x(0,0)}^0x-\overbrace{f_y(0,0)}^0y|}{||(x,y)||_2}=\dfrac{|x^3+y^3|}{x^2+y^2}\\\\\le\dfrac{x^2|x|+y^2|y|}{x^2+y^2}\le \dfrac{x^2+y^2}{x^2+y^2}\max(|x|,|y|)\end{align}$
The last part is obtained via $\max(|x|,|y|)=||(x,y)||_\infty\le||(x,y)||_2\to 0$
Best Answer
My answer shows it is not differentiable at origin. Kindly check whether I have done any calculation error. $${f_x} (0,0)=\mathop {\lim }\limits_{x \to 0} {{f(x,0) - f(0,0)} \over x} = \mathop {\lim }\limits_{x \to 0} {{x-0} \over x} = 1$$ $${f_y} (0,0)=\mathop {\lim }\limits_{y \to 0} {{f(0,y) - f(0,0)} \over y} = \mathop {\lim }\limits_{y \to 0} {{0 - 0} \over y}=0.$$ Therefore
$$I=\lim_{(x,y)\rightarrow (0,0) } {{f(x,y) - f_x(0,0)x-f_y(0,0)y-f(0,0)} \over{\sqrt{x^2+y^2}}} = \lim_{(x,y)\rightarrow (0,0) } {{{x^3/(x^2+y^2)}-x-0-0}\over {\sqrt{x^2+y^2}}}$$ $$=\lim_{(x,y)\rightarrow (0,0) } {-xy^2\over{(x^2+y^2)\sqrt{x^2+y^2}}}. $$ Calculate the limit value of the second using polar co-ordinates, by taking $x=r\cos\theta$ and $y=r\sin\theta$ and $r\rightarrow 0.$ Then $I=\lim_{r\rightarrow 0}\cos \theta\sin^2\theta =f(\theta).$ Limit value depends on the path. So limit does not exist and the function is not differentiable at the origin.