Find maximum value of $$f\left(x\right)=x\sqrt{x}-\frac{1}{\sqrt{1-x^2}}$$
Domain of the function is $[0,1)$. Equating the derivative by zero we get,
$$\frac32\sqrt x-\frac{x}{(1-x^2)^{\frac32}}=0$$ $$\frac94x=\frac{x^2}{(1-x^2)^3}$$ $$9(x^2-1)^3+4x=0$$
Not sure how to find the roots of the equation. But I can proceed as follow,
$$\frac9{4x}=\frac{1}{(1-x^2)^3}$$ $$\frac1{\sqrt{1-x^2}}=\sqrt[\large6]{\frac9{4x}}$$So we are looking for maximum of $x\sqrt x-\sqrt[\large6]{\frac{9}{4x}}$