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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 $x\sqrt x-\sqrt[\large6]{\frac{9}{4x}}$
You end with the problem of solving for $x$ the sextic equation $$9 x^6-27 x^4+27 x^2+4 x-9=0$$ and, by inspection (as @NickD already commented) the solution is close to $0.6$.
Make $x=y+\frac 35$ to make $$9 y^6+\frac{162 y^5}{5}+\frac{108 y^4}{5}-\frac{648 y^3}{25}-\frac{1728 y^2}{125}+\frac{53972 y}{3125}+\frac{636}{15625}=0$$ Neglect the terms $(y^3,\cdots,y^6)$ and solve the quadratic $$-\frac{1728 y^2}{125}+\frac{53972 y}{3125}+\frac{636}{15625}=0$$ The smallest root is $$y=-\frac{318}{5 \left(13493+\sqrt{183434809}\right)}\implies x=0.597647652\cdots$$ while the exact solution is $x=0.597647629\cdots$
Using this approximation, the maximum of the function is then $$-0.78523029529810916\cdots$$ while the exact solution is $$-0.78523029529810836\cdots$$ which is not too bad since obtained at the price of a quadratic equation.
