Let,

$y=\frac{x^3}{x^4+147}=f\left(x\right)$

We know that, for increasing function $\frac{dy}{dx}>0$

So, differentiating the given function $f\left(x\right)=\frac{x^3}{x^4+147}$ we get,

$\frac{dy}{dx}=\frac{-x^2\left(x^4-441\right)}{(x^4+147{)}^2}$

$\Rightarrow \frac{-x^2\left(x^4-441\right)}{(x^4+147{)}^2}>0$

$\Rightarrow \frac{x^2\left(x^4-441\right)}{(x^4+147{)}^2}<0$

$\Rightarrow x^4<441$

Now for maxima/minima $\frac{dy}{dx}=0$

$\Rightarrow x^4=441$

$\Rightarrow x=\alpha , 4<\alpha <5$

$\Rightarrow$Maximum value of $f\left(x\right)$ is at $x=4$ or $x=5$

$f\left(4\right)=\frac{64}{403}, f\left(5\right)=\frac{125}{772}$

$\because f\left(5\right)>f\left(4\right)$

$\Rightarrow \alpha =5$