Is my solution of $\lim_{(x, y)\rightarrow (0, 0)} \left( \frac{\cos (x^2+y)}{1-x^2y} \right) ^{\frac{1}{x^4+y^2}}$ correct? Is this solution correct?
If not, why?
Alternative (especially less wordy) proofs are also welcome.
$$\lim_{(x, y)\rightarrow (0, 0)} \left( \frac{\cos (x^2+y)}{1-x^2y} \right) ^{\frac{1}{x^4+y^2}}$$
$$y=x^2$$
$$\lim_{(x, x^2)\rightarrow (0, 0)}\left( \frac{\cos (2x^2)}{1-x^4}\right) ^{\frac{1}{2x^4}}$$
$$x^2=a$$
$$\lim_{a\rightarrow 0}\left( \frac{\cos (2a)}{1-a^2}\right) ^{\frac{1}{2a^2}}$$
$$\lim_{a\rightarrow 0} e^{\ln \left( \frac{\cos (2a)}{1-a^2}\right) ^{\frac{1}{2a^2}}}$$
Zooming in on the power:
$$\lim_{a\rightarrow 0} {\ln \left( \frac{\cos (2a)}{1-a^2}\right) {\frac{1}{2a^2}}}.$$
In the neighborhood of $0$, $\cos(2a)$ is barely distinguishable from $1$:
$$ \lim_{a\rightarrow 0} \frac{\ln\frac{1}{1-a^2}}{2a^2} = \langle \frac{0}{0} \rangle,$$
so L'Hôpital:
$$ \lim_{a\rightarrow 0} \frac{\frac{-(-2a)}{(1-a^2)^2}(1-a^2)}{4a} = \frac{1}{2} .$$
The answer: $\sqrt{e}.$

You have approached this by looking at one curve, namely $y=x^2.$ And that curve should have yielded the limit $e^{-1/2}.$ So there's that mistake. The bigger mistake is: One curve is never enough.
We can write
$$\tag 1 \cos (x^2+y) = 1-\frac{(x^2+y)^2}{2} +o(x^4+y^2)$$
and
$$\tag 2 \frac{1}{1-x^2y} = 1 + x^2y + o(x^4+y^2).$$
The product of $(1)$ and $(2)$ is then
$$\tag 3 1-\frac{x^4+y^2}{2}+ o(x^4+y^2).$$
Rasing $(3)$ to the power $\dfrac{1}{x^4+y^2}$ and taking the limit as $(x,y)\to (0,0)$ shows the desired limit is $e^{-1/2}.$
Note This is edited from the original, which contained a mistake that Peter Foreman pointed out to me.
