I think you have a typo in your $f(z)$. Either way, here is what you should do when $x = y$.
Consider this definition of the partial derivative: $$\frac{\partial f}{\partial x}(x_0,y_0) = \lim_{h\to 0} \frac{f(x_0 + h,y_0) - f(x_0,y_0)}{h}.$$
Set $x_0 = y_0$. $$\frac{\partial f}{\partial x}(x_0,x_0) = \lim_{h\to 0} \frac{f(x_0 + h,x_0) - f(x_0,x_0)}{h}.$$
Note that $f(x_0, x_0) = 0$. Therefore, $$\frac{\partial f}{\partial x}(x_0,x_0) = \lim_{h\to 0} \frac{f(x_0 + h,x_0)}{h} = \lim_{h\to 0}\; h^2\sin(1/h).$$
Notice how this value is the same for every $x_0$.
You are on the right track but the question is, how to compute the following in general :
$$\lim_{(x,y)\to(a,b)}h(x,y).$$
We may not just do $x \to a$ then $y \to b$ (or the converse) for example. We need to consider all the manners for $(x,y)$ to tend to $(a,b)$.
Example of why it does not work :
Consider $f(x,y)= \chi_{\{x=0\} \cup \{y=0\}}$. With $\chi_A$ the function whose value is $1$ on $A$ and $0$ otherwise. Then $f$ is not continuous at $0$ but :
$$\lim_{x \to 0} f(x,0)=1=\lim_{y \to 0} f(0,y).$$
But obviously the limit for both variables does not exists, since :
$$\lim_{t \to 0} f(t,t)=0.$$
Which brings us to one of the method to show a function is not continuous at a point. Just find two different manners to tend to the point of interest such that the limit is different.
In your case the answer is clear since the partial derivative you found are polynomials.
But one way (the most usual one I think) is to majorate the quantity $|h(a+u,b+v)-h(a,b)|$ by something who obviously tends to $0$ when $(u,v) \to 0$ or which had been proved to tend to $0$.
Finally, here is you want to prove that the function are continuous, i.e. if you consider not known that polynomials are continuous, you have to get back to definition of continuity with $\varepsilon, \delta$.
Let $\varepsilon >0$, suppose $\|(u,v)-(a,b)\|_1 \le \delta$ with $\delta = \varepsilon / 2$ then :
$$|f_x(u,v)-f_x(a,b)|=|2u+v-(2a+b)|\le 2|u-a| + |v-b| \le 2 \|(u,v)-(a,b)\|_1 \le \varepsilon$$
Then $f_x$ is continuous at $(a,b)$. I let you try for $f_y$.
Best Answer
We should find the value of partial derivatives at the point by the definition (via limits).
We have $f(x,0) = 0$ and $f(0,0) = 0$. Then, $$f'_x(0,0) = \lim_{x\to 0}\frac{f(x,0) - f(0,0)}{x-0} = 0$$ Similarly, we get $$f'_y(0,0) = \lim_{y\to 0}\frac{f(0,y) - f(0,0)}{y-0} = 0$$ Hence we need to check if the following equality is true: $$\lim_{x\to0 \\ y\to0} \frac{f(x,y) - f(0,0) - 0 - 0}{\sqrt{x^2 + y^2}} \overset ? = 0$$ which is equivalent to $$\lim_{x\to0 \\ y\to0} \frac{\frac{x^2y}{x^2+y^2}}{\sqrt{x^2+y^2}}= \lim_{x\to0 \\ y\to0} \frac{x^2y}{(x^2+y^2)^{3/2}} \overset ?= 0$$ Can you take it from here?