Make a graph showing all three curves. You’ll see that there is just one finite region bounded by these curves. It has two straight sides along the lines $y=x$ and $y=x/8$ and one curved side along $y=1/x^2$. If that third side ran straight from $\langle 1,1\rangle$ to $\left\langle 2,\frac14\right\rangle$, the region would be the triangle with vertices at the origin, at $\langle 1,1\rangle$, and at $\left\langle 2,\frac14\right\rangle$.
(In case it isn’t clear, $\langle 1,1\rangle$ is the point of intersection of $y=x$ and $y=1/x^2$, and $\left\langle 2,\frac14\right\rangle$ is the point of intersection of $y=x/8$ and $y=1/x^2$.)
You’ll want to set it up as two integrals, one from $x=0$ to $x=1$, the other from $x=1$ to $x=2$. Your vertical strips $dA$ of area for the first integral will run between the two straight lines; for the second integral they’ll run from the line $y=x/8$ to the curve $y=1/x^2$.