If $V$ is the volume of a cuboid, $S$ is its total surface area, and $a, b,\; \text{and}\; c$ are the areas of its sides, then we have to prove that
${1\over{V^2}}=\frac{2}{S}(\frac1{a}+\frac1b+\frac1c)$
I think that $S$ must be twice the sum of $a, b,\; \text{and}\; c$, but now what?
Let the length of edges be $x,y,z$.
The area of the sides are: $a=xy,b=yz,c=xz$
The volume is $V=xyz$
The TSA is $S=2(xy+yz+xz)$
Insert these into your equation.
${1\over{(xyz)^2}}=({2\over{2(xy+yz+xz)}})({{xy+yz+xz}\over{(xyz)^2}})$
Which simplifies to
${1\over{(xyz)^2}}={1\over{(xyz)^2}}$
Thus proved.