I need maximum of $$ \frac{x^{15}(1-x)y^{15}(1-y)}{(1-xy)^{15}}$$ where $0<x<1$, $0<y<1$
Define $$ f(x,y)=\frac{x^{15}(1-x)y^{15}(1-y)}{(1-xy)^{15}}$$ So the function is positive for $0<x<1$, $0<y<1$ and vanishes on the boundary $x=0$ or $x=1$ or $y=0$ or $y=1$. Therefore it attains the maximal value inside the region.
So for maxima or minima $\frac{\partial{f}}{\partial{x}}=0$ and $\frac{\partial{f}}{\partial{y}}=0$
Using Wolfram Alpha $$\frac{\partial{f}}{\partial{x}}=-\frac{x^{14}(y-1)y^{15}(x^2y-16x+15)}{(1-xy)^{16}}=0$$ Which gives $$x^2y-16x+15=0\tag{1}$$ Similarly $\frac{\partial{f}}{\partial{y}}=0$ gives $$y^2x-16y+15=0\tag{2}$$ Subtracting $(1)$ and $(2)$, we obtain $$ x^2y-y^2x+16y-16x=0 $$ So we get $$(xy-16)(x-y)=0 $$ which gives $x=y$ or $xy=16$. Since $0<x,y<1$, so $xy=16$ is not possible. So we have for maxima or minima $$x=y\tag{3}$$ So we are down to function of single variable $$ F(x)=\frac{x^{30}(1-x)^2}{(1-x^2)^{15}}$$ where $0<x<1$. Any help will be highly appreciated. Thank you
Consider the limit of $f(x,y)$ as $(x,y)$ approaches $(1,1)$ along the path $x=y$. Then $$\lim_{(x,y)\to(1,1)}f(x,y)=\lim_{x\to1}\frac{x^{30}(1-x)^2}{(1-x^2)^{15}}=\lim_{x\to1}\frac{x^{30}}{(1-x)^{13}(1+x)^{15}}=\infty.$$ This shows that $f(x,y)$ does not have a global maximum in the given domain.
At any local extremum you have $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$. You have already shown that this implies $x=y$. Then also $\frac{d}{dx}f(x,x)=0$ where $$\frac{d}{dx}f(x,x)=\frac{x^{29}(1-x)^2(x^2+x-15)}{(1-x^2)^{16}},$$ but the numerator is easily verified to be nonzero for $0<x<1$. So $f(x,y)$ does not even have any local extremum in the given domain.