$\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{\sin(x)\cos(y)}$ is this done correctly?

$\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{\sin(x)\cos(y)}$ is it allowed to split a multi-variable limit into its component variables as in the next step?

$= (\lim_{x\to0}\frac{x^2}{\sin(x)})(\lim_{x\to0}\frac{y^2}{\sin(y)})$ this is an indeterminate form and now I use L'Hopital

$=(\lim_{x\to0}\frac{2x}{\cos(x)})(\lim_{x\to0}\frac{2y}{\cos(y)})$

$=(\frac{0}{1})(\frac{0}{1})=0$

Even switching the order of taking a limit can lead to problems.

As an example: $\lim\limits_{m \to \infty} \lim\limits_{n \to \infty} {\frac{1}{n}} ^{\frac{1}{m}} = \lim\limits_{m \to \infty} 0 ^{\frac{1}{m}} = 0 \neq 1 = \lim\limits_{n \to \infty} 1 = \lim\limits_{n \to \infty} \lim\limits_{m \to \infty} \frac{1}{n} ^\frac{1}{m}$

While: $\lim\limits_{(m,n) \to (\infty, \infty)} {\frac{1}{n}} ^{\frac{1}{m}}$ does not exist.

At this point I would like to refer to this post well written post and hope this answers some further questions of yours: https://math.stackexchange.com/q/15257

No, it is generally not allowed as it may be false. But you can do as follows:

$$\lim_{(x,y)\to(0,0)}\;x\cdot\frac x{\sin x}\cdot\frac{y^2}{\cos y}=0\cdot1\cdot\frac01=0$$

It is actually easier: $$\left|\frac{x^2 y^2}{\sin x \cos y}\right| \leq 2\left|\frac{x}{\sin x} \right| \left| y^2 \right| \left| x \right| \leq |xy^2| =o(1)$$ as $(x,y) \to (0,0)$.