Let $f \, : \, \mathbb{R}^{d} \times \mathcal{M}_{d}(\mathbb{R}) \times \mathbb{R}^{d} \, \longrightarrow \, \mathbb{R}$ such that :
$$ f(x,V,y) = x^{\top}Vy $$
$\nabla_{x}f$ (resp. $\nabla_{V}f$, resp. $\nabla_{y}f$) denote the gradient of $f$ with respect to the variable $x$ (resp. $V$, resp. $y$), in other words, the gradient of the linear map $x \, \longmapsto \, f(x,V,y)$ (resp. $V \, \longmapsto \, f(x,V,y)$, resp. $y \, \longmapsto \, f(x,V,y)$). We have :
$$
\left\{
\begin{array}{l}
\nabla_{x}f \, (x,V,y) = Vy \\[2mm]
\nabla_{V}f \, (x,V,y) = xy^{\top} \\[2mm]
\nabla_{y}f \, (x,V,y) = V^{\top}x \\
\end{array}
\right.
$$
-- Update :
Let $(E,\left\langle \cdot,\cdot \right\rangle)$ be an euclidean space of dimension $n$. Let $f \, : \, E \, \longrightarrow \, \mathbb{R}$ and a point $x \in E$. Let $\mathcal{L}(E,\mathbb{R})$ the space of [continuous] linear forms on $E$. By definition, $f$ is differentiable at $x$ if there exist a linear form $ \in \mathcal{L}(E,\mathbb{R})$ such that :
$$ f(x+h) = f(x) + L(h) + \mathop{o} \limits_{\Vert h \Vert \to 0}\big( \Vert h \Vert \big)$$
$L$ is unique and I will use the notation : $L=\mathrm{D}_{x}f$ ( $\mathrm{D}_{x}f$ is the differential of $f$ at $x$) and $L(h) = \mathrm{D}_{x}f \cdot h$. Since $\mathrm{D}_{x}f$ is in $\mathcal{L}(E,\mathbb{R})$, Riesz representation theorem ensures that there exist a unique $x_{f} \in E$ such that : $\forall h, \, \mathrm{D}_{x}f \cdot h = \left\langle h,x_{f} \right\rangle$ (one could say that the linear form $\mathrm{D}_{x}f$ is "represented" by the vector $x_{f}$ throught the inner product $\left\langle \cdot,\cdot \right\rangle$). Since $x_{f}$ is unique, we say that the vector $x_{f}$ is the gradient of $f$ at $x$ and we write usually $x_{f} = \nabla f(x)$.
If $E = \mathbb{R}^{n}$ and $\left\langle \cdot,\cdot \right\rangle$ is the usual inner product, then : $\mathrm{D}_{x}f \cdot h = \left\langle h,\nabla f (x) \right\rangle = h^{\top} \nabla f (x)$.
If $E = \mathcal{M}_{d}(\mathbb{R})$ (the space of real square matrices of size $d$) and $\left\langle A,B \right\rangle = \mathrm{Tr}(A^{\top}B)$, then : $D_{x}f \cdot h = \left\langle h,\nabla f(x) \right\rangle = \mathrm{Tr}(h^{\top} \nabla f(x))$. (Note that in this case, $\nabla f(x)$ is a matrix, since it is an element of $E$).
So, how could we proceed in order to get the gradient of $\varphi \, : \, x \, \longmapsto \, x^{\top}Vy$ ($V$ and $y$ are fixed here) ? Write :
$$
\begin{align*}
\varphi(x+h) &= {} \big( x+h \big)^{\top}Vy \\[2mm]
&= x^{\top}Vy + h^{\top}Vy \\
\end{align*}
$$
In $\varphi(x+h)$, identify the $\varphi(x)$ part and the part which is linear in $h$, the differential of $\varphi$ at $x$ :
$$ \varphi(x+h) = \underbrace{x^{\top}Vy}_{\varphi(x)} + \underbrace{\color{red}{h^{\top}Vy}}_{\mathrm{D}_{x}\varphi \, \cdot \, h} $$
Now, we know that the differential and the gradient are linked by the relation : $\mathrm{D}_{x}\varphi \cdot h = h^{\top} \nabla \varphi (x)$ so we identify $h^{\top}Vy$ and $h^{\top} \nabla \varphi (x)$ :
$$ \nabla \varphi (x) = Vy $$
We can do the same for $\psi \, : \, V \, \longmapsto \, x^{\top}Vy$ :
$$
\begin{align*}
\psi(V+H) &= {} x^{\top}\big( V+H \big)y \\[3mm]
&= \underbrace{x^{\top}Vy}_{\psi(V)} + \underbrace{\color{red}{x^{\top}Hy}}_{\mathrm{D}_{V} \psi \, \cdot \, H} \\
\end{align*}
$$
Here, $\mathrm{D}_{V}\psi \cdot H$ is not yet of the form $\mathrm{Tr}(H^{\top} \nabla \psi(V))$ so let's write :
$$
\begin{align*}
x^{\top}Hy &= {} \mathrm{Tr}\big( x^{\top}Hy \big) \\[2mm]
&= \mathrm{Tr}\big( yx^{\top}H \big) \\[2mm]
&= \mathrm{Tr}\big( H^{\top}xy^{\top} \big) \\
\end{align*}
$$
As a consequence :
$$ \nabla \psi(V) = xy^{\top}. $$
Best Answer
Let $Y=1^T\Phi$, then the problem is to find the derivative of the function $\,L=\|Y\|_F^2$
Better yet, using the Frobenius product, the function can be written as $\,L=Y:Y$
Start by taking the differential $$\eqalign{ dL &= 2\,Y:dY \cr &= 2\,1^T\Phi:1^Td\Phi \cr &= 2\,11^T\Phi:d\Phi \cr &= 2\,11^T\Phi:\Phi'\circ d(XV) \cr &= 2\,(11^T\Phi)\circ\Phi':d(XV) \cr &= 2\,(11^T\Phi)\circ\Phi':X\,dV \cr &= 2\,X^T[(11^T\Phi)\circ\Phi']:dV \cr }$$ Since $dL = \big(\frac {\partial L} {\partial V}\big):dV\,\,$ the derivative must be $$\eqalign{ \frac {\partial L} {\partial V} &= 2\,X^T[(11^T\Phi)\circ\Phi'] \cr }$$ This is the same result as @legomygrego, but with the step-by-step details. The only property which might be new to some readers is the mutual commutivity of the Frobenius and Hadamard products $$\eqalign{ A:B &= B:A \cr A\circ B &= B\circ A \cr A\circ B:C &= A:B\circ C \cr }$$