The point at which the tangent line to the curve of $y=\frac{16}{x}-x^2$ is horizontal

Given equation of curve is $y=\frac{16}{x}-x^2...............\left(i\right)$

Differentiating w. r. t. $x$

$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\frac{16}{x}-x^2\right)$

$\frac{dy}{dx}=-\frac{16}{x^2}-2x$

The slope of tangent, $\frac{dy}{dx}=0$

$-\frac{16}{x^2}-2x=0$

$-\left(\frac{16+2x^3}{x^2}\right)=0$

$x^3=-8$

$x=-2$

Putting value $x$ in equation $\left(i\right)$

$y=\frac{16}{-2}-{\left(-2\right)}^2$

$y=-12$

Hence, the point is $\left(-2,-12\right)$.