We define
$$\displaystyle f(x,y)=\frac{1}{x^{2y}-\frac{1}{4^y}}+\frac{1}{(1-x)^{2y}-\frac{1}{4^y}} \text{ for } (x,y) \in \left[0,\frac{1}{2}\right) \times \left(\frac{1}{2},1\right]$$.
A study with Maple suggests that
$$f(x,y) \geq f(0,1)=-\frac{8}{3} \text{ for } (x,y) \in \left[0,\frac{1}{2}\right) \times \left(\frac{1}{2},1\right]$$
Is someone able to prove this analytically? The derivatives, unfortunately, don't look very nice.
I'm glad about every idea!
For a fixed $x$, the derivative of $f(x,y)$ is given by $$\frac{\partial f}{\partial y}=-\frac{4^y(\ln4+4^y(1-x)^{2y}\ln(1-x)^2}{(4^y(1-x)^{2y}-1)^2} - \frac{4^y(\ln4+4^y (x^{2y}\ln x^2)}{(4^y x^{2y}-1)^2}<0$$ Hence $f$ is decreasing with increasing $y$, and so takes its minimum value for $y=1$.
$$f(x,1)=\frac{1}{x^2-4^{-1}}+\frac{1}{(1-x)^2-4^{-1}}$$
$$\frac{\partial f}{\partial x}(x,1)=32\times\frac{1-2x}{(3+4x-4x^2)^2}$$ It is easy to check that this fraction is positive in the given range, so the minimum is achieved at the smallest allowed value of $x$, namely $0$.
Edit: To prove the first inequality, first simplify it via $a:=e^y$, $t:=-\log_4(4x^2)$, $s=-\log_44(1-x)^2$, $$-\frac{1}{4^y\ln4}\frac{\partial f}{\partial y}=\frac{1-a^s(1-s)}{(a^s-1)^2}+\frac{1-a^t(1-t)}{(a^t-1)^2}$$ This can be shown by splitting the ranges of $s$ and $t$.