I would suggest starting by finding where they intersect! You are given the curves \(\displaystyle y= Cx^2\) and \(\displaystyle y= 1/x^2\). From \(\displaystyle y= Cx^2\), \(\displaystyle x^2= y/C\) so that \(\displaystyle y= 1/x^2= C/y\). So \(\displaystyle y^2=C\) so that \(\displaystyle y= \sqrt{C}\) or \(\displaystyle y= -\sqrt{C}\). y cannot be \(\displaystyle -\sqrt{C}\) because we cannot have \(\displaystyle y= -\sqrt{C}= 1/x^2\). So we have \(\displaystyle y= \sqrt{C}= 1/x^2\) so that \(\displaystyle x^2= \frac{1}{\sqrt{C}}\) and so \(\displaystyle x= \frac{1}{\sqrt[4]{C}}\) or \(\displaystyle x= -\frac{1}{\sqrt[4]{C}}\).
The two points of intersection are \(\displaystyle \left(\frac{1}{\sqrt[4]{C}}, \sqrt{C}\right)\) and \(\displaystyle \left(-\frac{1}{\sqrt[4]{C}}, \sqrt{C}\right)\).
(Equivalently, since \(\displaystyle y= \frac{1}{x^2}\), \(\displaystyle y= Cx^2\) becomes \(\displaystyle Cx^2= \frac{1}{x^2}\). Multiplying on both sides by \(\displaystyle \frac{x^2}{C}\), \(\displaystyle x^4= \frac{1}{C}\) so that \(\displaystyle x= \frac{1}{\sqrt[4]{C}}\) as before.)
Now those two curves will be "orthogonal" when the tangent lines at those points are perpendicular. And that will occur when the product of the slopes is -1. Of course, the slopes of the tangent lines are the values of the derivatives at the point.
So find the derivative of each of these functions, at the points of intersection (I suspect, because of symmetry, that either one will do) and find C so that the product of the derivatives is -1.
