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

The given curve is $y=x+\frac{4}{x^2}$

Differentiate both sides 

$\frac{dy}{dx}=1-\frac{8}{x^3}$

Since the tangent is parallel to $x-$ axis, slope of the tangent is zero

$\begin{array}{rcl}1-\frac{8}{x^3}& =& 0\\ \frac{8}{x^3}& =& 1\\ x^3& =& 8\\ x& =& \sqrt[3]{8}\\ & =& 2\end{array}$

Equation of the tangent which is parallel to $x-$ axis to the curve is $y=1+\frac{4}{4}=1+1=2\Rightarrow y=\boxed{\mathbf{2}}$