Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$
$$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=\pm \sqrt{\frac{2}{3}}$
But $x \ne 0$ since $xy=-1$
$$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$
Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$
Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$
Is there any other approach? 

With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=\frac4{4-t}+\frac{9t}{9t-1},t\in\left(\frac14,4\right)$$
Then
$$h(t)=\frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$\log h(t)=\log(-9t^2+72t-4)-\log(4-t)-\log(9t-1)$$
$$(\log h(t))'=\frac{-18t+72}{-9t^2+72t-4}+\frac1{4-t}-\frac9{9t-1}=\frac{315t^2-140}{\text{Irrelevant}}$$
This is $0$ for $t=\frac23$
