Evaluate $\lim_{(x,y)\to(0,0)}\frac{\cos(xy)-1}{x^2y^2}$

You know that $$\lim_{t\to 0} \frac{\cos t-1}{t^2} =-\frac{1}{2}, $$using L'Hospital, for example. This tells us that $g\colon \Bbb R \to \Bbb R$ defined by $$g(t) \doteq \begin{cases} \dfrac{\cos t-1}{t^2}, \mbox{if } t \neq 0 \\[.5em] -\dfrac{1}{2}, \mbox{if }t=0\end{cases}$$is continuous. Because of that, we can write $$\lim_{(x,y)\to (0,0)} \frac{\cos (xy)-1}{x^2y^2} = \lim_{(x,y) \to (0,0)} g(xy) = g\left(\lim_{(x,y)\to (0,0)}xy\right) = g(0) = -\frac{1}{2}.$$

You are basically done,

I would break the expression inside the limit into,

$$\frac{-1}{1+\cos t} \frac{\sin t}{t} \frac{\sin t}{t}$$

Then use the product rule for limits and the famous limit $\frac{\sin t}{t} \to 1$ as $t \to 0$.

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Edit

As per the answer of @Steven Stadnicki, your limit is only equal to the single variable limit if $xy \neq 0$ as we approach the origin. Otherwise, the expression is not even defined even if we are very close to the origin. It's safer to say that $f(x,y)=\frac{\cos (xy)-1}{(xy)^2}$ tends to negative a half as we approach the origin going through points which are a part of the domain of $f$ in which the function is defined.

Hint. One may recall that, as $t \to 0$, $$ \cos t \to 1,\qquad \frac{\sin t}{t} \to 1, $$ giving, as $t \to 0$, $$ \dfrac{-\sin^{2}t}{t^2(\cos(t)+1)}\to -\frac12. $$