The points on the curve $y=x^4-4x^3+4x^2+1$ at which the tangents are parallel to x-axis are

$y=x^4-4x^3+4x^2+1$

Slope$={\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=4x_1^3-12x_1^2+8x_1$    

The points which tangent are  parallel to x-axis are  ${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=0$

$\Rightarrow 4x_1^3-12x_1^2+8x_1=0$

$\Rightarrow 4x_1\left(x_1^2-3x_1+2\right)=0$

$\begin{array}{c}\therefore x_1=0,1,2\\ y_1=1,2,1\end{array}\}\Rightarrow \left(0,1\right),\left(1,2\right),\left(2,1\right)$