C'mon it's quite easy

Since the line $y = mx +1$ touches the curve $y=-x^4+2x^2+x$ at $P(x_1, y_1)$ and $Q(x_2, y_2)$ (see above), at these two points the following equations are satisfied.

$\begin{cases} y =-x^4+2x^2+x = mx + 1 & ...(1) \\ \dfrac {dy}{dx} = -4x^3+4x+1 = m & ...(2) \end{cases}$

$Eq1-Eq.2\times x: \quad \Rightarrow 3x^4-2x^2 = 1 \quad \Rightarrow (3x^2+1)(x^2-1)=0$

$\Rightarrow x = \pm 1\Rightarrow \begin{cases} x_1=1 & \Rightarrow y_1 = -1+2+1 = 2 \\ x_2=-1 & \Rightarrow y_1 = -1+2-1 = 0 \end{cases}$

$\Rightarrow x_1^2 + x_2^2+y_1^2 + y_2^2 = 1+1+4+0 = \boxed{6}$

According to the question we can say that :

$-x^4+2x^2+x=mx+1$ has solutions $x_1,x_1$ and $x_2,x_2$ .

$\implies x^4-2x^2+(m-1)x+1=(x-x_1)^2 \cdot (x-x_2)^2$

On comparing coefficients, we get : $x_1=1,x_2=-1$ and $m=1$

Putting values of $x_1,x_2$ in the equation of curve, we get $y_1=2,y_2=0$ .

Hence $\boxed{x_1^2+x_2^2+y_1^2+y_2^2=6}$ .

enjoy !

The point of intersection of two curves is given by equating the line and curve. So , ${ x }^{ 4 }-2{ x }^{ 2 }+(m-1)x+1=0$ At these points of intersection the slopes of the two curves are equal as well, So $4{ x }^{ 3 }-4x+m-1=0$ The two equations must have two common solutions. $x=+1,-1$ satisfies them hence the answer

m=1 ; x= 1,-1