Note that $f(0,t) = 0$ (for $t \neq 0$), but $$f(t,-t+t^2) = \frac{t^2+t^2-t^3}{t^2} = 2-t,$$ again for $t \neq 0$. What happens when $t \to 0$? This shows that $f$ is not continuous at $(0,0)$.
On the other hand, $f$ is continuous everywhere else where $f$ is defined, since the numerator and denominator clearly are continuous.
Although your argument contains a grain of truth, it is not quite correct as it is written, since you wrote that the limit $\lim_{(x,y)\to (0,0)} f(x,y)$, which doesn't exist, is equal to the limit $\lim_{r \to 0} (\cdots)$, which does exist (for $\sin \theta \neq 0$) if you just view it as an ordinary single-variable limit which depends parametrically on a constant $\theta$; in fact, you just computed it yourself like that and wrote that it's equal to $1 + \cot \theta$.
The problem is that you want $\theta$ to be able to vary independently of $r$ as $r \to 0$, so you can't treat $\theta$ as a constant here. It should be viewed as an arbitrary function $\theta(r)$.
In fact, polar coordinates are mainly useful for proving that a limit exist, namely if you can write $f(x,y)$ as a bounded factor times another factor which depends only on $r$ (no $\theta$!) and which tends to zero as $r \to 0$ (really just a single-variable limit here!), then $f(x,y)\to 0$ as $(x,y)\to(0,0)$.
To show that a limit does not exist, you instead find two ways of approaching the point such that you get two different values. In your case, consider $f(t,0)$ and $f(0,t)$ as $t\to 0$, for example.
So actually I don't quite know how I would like to write the argument in a nice way if someone forced me to use polar coordinates in order to show that a limit does not exist! I would probably just write $f(x,y)$ in polar coordinates first,
$$
f(r \cos\theta(r), r \sin\theta(r)) = \cdots,
$$
(no “$\lim$” here) and then say that by making different choices of the function $\theta(r)$ (for example different constant functions!), you can make $f$ approach different values. And I would give examples of two such function $\theta(r)$ which give different limits for $f(r \cos\theta(r), r \sin\theta(r))$ as $r \to 0$.
Best Answer
You don't need to change to polar coordinates. Notice that at any point $(x,y)\ne (0,0)$ the function is continuous since it is a quotient of continuous functions. So, you just need to check the limit as $(x,y)$ goes to $(0,0)$. Noting that as $(x,y)$ goes to $(0,0)$ so does $x^2+y^2$, and since $\lim_{t\to 0}\sin(t)/t=1$ you may conclude that the limit of the function as $(x,y)\to (0,0)$ is equal to $0$ (note the multiplication by $x$ in the numerator). Since $f(0,0)=0$ be definition it follows that the function is continuous everywhere.