Not quite.
$$
\frac{\partial}{\partial x}=\frac{\partial}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}
$$
However, $r=\sqrt{x^2+y^2}$, so $\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\cos (\theta )$. Similarly, $x=r\cos (\theta )$, so $1=\frac{\partial r}{\partial x}\cos (\theta )-r\sin (\theta )\frac{\partial \theta}{\partial x}$, so that $\frac{\partial \theta}{\partial x}=-\frac{\sin (\theta )}{r}$. Plugging this in above, we find
$$
\frac{\partial}{\partial x}=\cos (\theta)\frac{\partial}{\partial r}-\frac{\sin (\theta )}{r}\frac{\partial}{\partial \theta}
$$
From here, you have to differentiate this with respect to $x$ to obtain an expression in polar coordiantes for $\frac{\partial ^2}{\partial x^2}$ (you essentially do the same thing I just did, although I'm sure it's a lot more tedious). And then you have to do the same for $y$. It is quite tedious. In the end, you wind up with
$$
\frac{\partial ^2}{\partial x^2}+\frac{\partial ^2}{\partial y^2}=\frac{1}{r}\frac{\partial}{\partial r}\left[ r\frac{\partial}{\partial r}\right] +\frac{1}{r^2}\frac{\partial ^2}{\partial \theta ^2}.
$$
I wouldn't worry too much about memorizing the formula. You should know how to derive it if need be and where to look it up if you need to (Wikipedia would surely have it for example).
If you need to see more details about the calculation, let me know.
-Jonny Gleason