Finding $ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $ $$ \lim_{(x,y) \to (0,0)} \frac{\sin^2(xy)}{3x^2+2y^2} $$
If I pick $ x = 0$ I get:
$$ \lim_{(x,y) \to (0,0)} \frac{0}{2y^2} = 0$$
So if the limit exists it must be $0$
Now for ${(x,y) \to (0,0)}$ I have $xy \to 0$
So I can use the Taylor series of $sin(t) = t + o(t)$ where $t \to 0$
$$0 \leq \frac{\sin^2(xy)}{3x^2+2y^2} = \frac{x^2y^2 + 2o(x^2y^2) + o (x^2y^2)}{3x^2 + 2y^2} =$$
$$\frac{x^2y^2 + o (x^2y^2)}{3x^2 + 2y^2} = $$
Now I can use the polar coordinates:
$$\frac{\rho^4 \cos^2(\theta)\sin^2(\theta) + o (\rho^4 \cos^2(\theta)\sin^2(\theta))}{3\rho^2 \cos^2(\theta) + 2\rho^2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 \cos^2(\theta) + 2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - 3 \sin^2(\theta) + 2 \sin^2(\theta)}=$$
$$\frac{\rho^2 \cos^2(\theta)\sin^2(\theta)(1 + o (1))}{3 - \sin^2(\theta)}\leq$$
$\frac{\rho^2}{2} \to 0$ for $\rho \to 0$
I would like to know if it is solved in the right way

We have $\;3x^2+2y^2\geq 2\sqrt{6}|xy|\;$ by AM-GM, thus we have
$\dfrac{\sin^2(xy)}{2\sqrt{6}|xy|}=\left(\dfrac{\sin(xy)}{xy}\right)^2\dfrac{|xy|}{2\sqrt{6}}\to0\;$ for $\;(x,y)\to(0,0)\;.$
I think your proof is ok. You can read also my proof if you want, it is a little bit shorter.
