What is the derivative of this? I have a function of the following form:
$J = \|W^TW-I\|_F^2$ 
Where, $W$ is a matrix and $F$ is the Frobenius Norm.
How can I find the derivative of $\frac{\partial J}{\partial W}$ ?

Assuming the norm comes from real inner-product. The derivative of the norm can be found as
$$\left.\frac{d}{dt}\right|_{0}\|(W+tH)^T(W+tH)-I\|^{2}\\
=\left.\frac{d}{dt}\right|_{0}\langle (W+tH)^T(W+tH)-I,(W+tH)^T(W+tH)-I\rangle\\
=2\langle W^TW-I,H^TW+W^TH\rangle$$
We now use the definition of adjoint corresponding to a simple matrix transpose for real matrices, $\langle Ax,y\rangle=\langle x,A^Ty\rangle$ and similarly for the operators acting from the right; as well as again the symmetry of the real inner product, we find that the two H-linear terms are in fact the same:
$$
\langle W^TW-I,W^TH\rangle=\langle W(W^TW-I),H\rangle
$$
$$
\langle W^TW-I,H^TW\rangle=\langle H(W^TW-I),W\rangle\\=\langle H,W(W^TW-I)^T\rangle\\
=\langle H,W(W^TW-I)\rangle\\
=\langle W(W^TW-I),H\rangle
$$
Summing up, the derivative is the linear map
$$
4\langle W(W^TW-I),H\rangle
$$
Then, and only because we have the inner product, we can write the gradient vector: $$\frac{\partial J}{\partial W}=4W(W^TW-I)$$
