Let $f:\mathbb{R}^{m\times n} \rightarrow \mathbb{R}^{m \times n}$ be a differentiable function. I want to take the gradient of the Frobenius inner product between $X$ and $f(X)$ with respect to $X$. That is, I want to differentiate the scalar function \begin{align} \phi(X)&= X : f(X)= \text{tr}(X^T f(X)) \end{align} with respect to $X$. I think the product rule applies (right?), so we get \begin{align} \partial \phi(X) &= \partial X : f(X) + X : \partial f(X). \end{align} I have the following questions:
If $f$ happens to satisfy \begin{align} \partial f(X):X = f(X):\partial X, \tag{$*$}\label{eq:1} \end{align} am I allowed to write \begin{align}\frac{\partial \phi(X)}{\partial X} = 2 f(X) \tag{$**$}\label{eq:2} \ ? \end{align}
We say that $f$ is homogeneous if $f(cX)=cf(X)$ for all $c\ge 0$. If $n=1$, then $f$ can be seen as a mapping from $\mathbb{R}^m$ to $\mathbb{R}^m$. In this particular case, the homogeneity condition implies \eqref{eq:1} (right?). Does it also work for $n>1$?
EDIT: In the vector case $(n=1)$, the directional derivative of $f: \mathbb{R}^m \rightarrow \mathbb{R}^m$ with respect to $x \in \mathbb{R}^m$ and also evaluated at $x$ is given by \begin{align} \frac{\partial f}{\partial x} \cdot x &= \lim_{\epsilon \rightarrow 0} \frac{f(x+\epsilon x) - f(x)}{\epsilon} \\ &= \lim_{\epsilon \rightarrow 0} \frac{(1+\epsilon) f(x) - f(x)}{\epsilon} \\ &= f(x), \end{align} and I believe \eqref{eq:1} must follow from that, but I'm struggling to make this point precise too.