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

$y=x+\frac{4}{x^2}\Rightarrow \frac{dy}{dx}=1-\frac{8}{x^3}$

Equation of tangent is parallel to x-axis 

$\therefore \frac{dy}{dx}=0\Rightarrow 1-\frac{8}{x^3}=0\Rightarrow x=2$

at $x=2,y=2+\frac{4}{4}=3\Rightarrow y_{ }=3$

$\therefore$ Equation of tangent is 

$y-y_1=0\left(x-x_1\right)\Rightarrow y=3$