I'd like to calculate the following derivative,
\begin{equation} \frac{d\|(f(C)\cdot f(C)^{+}-I)\cdot u\|^2)}{dC} \end{equation}
Where $C$ is a matrix of dimension $n\times k$ (s.t $k < n$). And $f$ is a differentiable function of $C$. Do you have suggestions on how to solve this problem using MATH tools or a cookbook?
Write the function in terms of the Frobenius product (:) and find its differential $$\eqalign{ L &= (ff^+-I)U:(ff^+-I)U \cr\cr dL &= 2\,(ff^+-I)U:d(ff^+)U \cr &= 2\,(ff^+-I)UU^T:(df\,f^+ + f\,df^+) \cr &\equiv W:(df\,f^+ + f\,df^+) \cr &= Wf^{+T}:df \,+\, f^TW:df^+ \cr }$$ Now the trick is to know that $f^T(ff^+-I)=0$ and therefore $f^TW=0$.
This reduces the differential to $$\eqalign{ dL &= Wf^{+T}:df \cr }$$ Since $dL = (\frac{\partial L}{\partial f}):df$ the derivative must be $$\eqalign{ \frac{\partial L}{\partial f} &= Wf^{+T} \cr &= 2\,(ff^+-I)UU^T f^{+T} }$$ To get the derivative with respect to $C$ I will make a few assumptions, since you didn't tell us anything about the function $f(C)$.
Since the shape of $C$ is not square, I assume that you are applying a scalar function element-wise to the matrix components. And that you know the derivative of the function in the scalar case, i.e. $f'(x)= \frac{df(x)}{dx}\,\,\,$
When applied element-wise to a matrix argument ($C$), the differential of a function can be expressed using the Hadamard product ($\circ$) as $$\eqalign{ df &= f' \circ dC \cr }$$ The differential of $L$ can be written as $$\eqalign{ dL &= \Big(\frac{\partial L}{\partial f}\Big):df \cr &= \Big(\frac{\partial L}{\partial f}\Big):(f' \circ dC) \cr &= \Big(\frac{\partial L}{\partial f}\circ f'\Big): dC \cr }$$ That means the derivative of $L$ with respect to $C$ is $$\eqalign{ \frac{\partial L}{\partial C} &= \Big(\frac{\partial L}{\partial f}\Big)\circ f' \cr }$$ If you are uncomfortable with the Frobenius product, you can replace it with the trace function, since $\,\,A:B = {\rm tr}(A^TB)$.