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 $x_0$.
Why are we allowed to find partial derivatives by holding a variable constant?
This can be seen directly from the definition:
$$\frac{\partial f(x,y)}{\partial x} = \lim\limits_{h\to0} \frac{f(x+h,y)-f(x,y)}{h}$$
In order to compute the limit, we can assume $y$ to be a parameter inside the limit. To make this more intuitive, we are going to let $g(x)=f(x,y)$. Applying this, we get
$$\frac{\partial f(x,y)}{\partial x} = \lim\limits_{h\to0} \frac{g(x+h)-g(x)}{h}$$
and this is just the derivative of a one-variable function. Additionally, when you are looking for derivatives the "easy way", you use various rules. But these rules sometimes come with some caveats, that we usually forget about, because most problems we solve are "nice enough" and do not require any special thought.
Why can you not use the method?
In this specific example, you would use the quotient rule, but, as any rule, it is only valid when the function is defined. This is clearly not the case for $(x,y)=(0,0)$ because the denominator is zero. So, the partial derivative of $\frac{x^3 + x^4 - y^3}{x^2+y^2}$ is undefined at $(0,0)$, and you cannot use the derivative of this function to find the derivative of $f$. But, this is not enough to say that $f$ has no partial derivatives at $(0,0)$, because it is defined there.
But, even if $f$ were defined at $(0,0)$, you would still be unable to use the method. Consider the following function:
$$f(x,y) = \begin{cases} x, & x\ne0 \\ 1, & x=0 \end{cases}$$
The derivative of $x$ is always $1$, but the partial derivative of $f$ with respect to $y$ doesn't exist at $0$. An easy way to see this is to notice that the function is not continuous at that point.
Best Answer
Yes, your solution is correct.
You may want to observe that for a fixed value $y\notin \{-1,0,1\}$ the function $x\mapsto f(x,y)$ is not even continuous at $0$, so it can't be differentiable.