EDIT: The question specifies that $f(0,0)=0$ in a piecewise.
For the first few parts of a question, I have used the definition of partial derivatives to show that $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ both exist at the point $(0,0)$ and are equal to $0$. I have also found the partial derivatives for $f$ at a point away from $(0,0)$ using normal differentiation rules like product, quotient and chain rule, and have shown that $\frac{\partial{f}}{\partial{x}}$ is not continuous at $(0,0)$ by showing its limit diverges as we approach $(0,0)$.
Have I made a mistake so far in my reasoning and working? If not, I would like to take on the last part of this question, which asks us to determine if $f$ is differentiable at $(0,0)$.
For this last part, I know two theorems:
1: "If $f$ is differentiable at some point, then all of its partial derivatives exist at that point and its derivative is the jacobian matrix of the function".
2: "If all of the partial derivatives of $f$ exists and are continuous on some set, then $f$ is differentiable on that set"
Could I use any of these here?
I know we could also use the definition of differentiability to calculuate the limit of and prove the existence of an affine approximation, but I was just wondering if there was any other way to do the question.
Since you already showed that $\frac{\partial{f}}{\partial{x}}$ and $\frac{\partial{f}}{\partial{y}}$ both exist at the point $(0,0)$ and are equal to $0$, your theorem 1 indicates that if $f$ is differentiable at this point, its differential is necessarily the zero map.
Therefore, by definition of the differential, $f$ is differentiable at $(0,0)$ if and only if $$\lim_{(x,y)\to(0,0)}\frac{|f(x,y)-f(0,0)-0x-0y|}{\|(x,y)\|}=0,$$ i.e. $$\lim_{r\to0}\frac{r^2\sin(1/r^2)}r=0,$$ which is indeed the case since $|\sin|\le1$.