The equation of the tangent to the curve $y=x+\frac{4}{x^2}$, that is parallel to the $x$ -axis, is:

A tangent parallel to the $x$ -axis has slope $\frac{dy}{dx}=0$.$y=x+4x^{-2}$$\frac{dy}{dx}=1-8x^{-3}=1-\frac{8}{x^3}$ Setting $1-\frac{8}{x^3}=0⟹x^3=8⟹x=2$.At $x=2$, $y=2+\frac{4}{2^2}=2+1=3$.The point is $(2,3)$. The equation of the line passing through $(2,3)$ with slope $0$ is $y=3$.