Let $E_i$ be a $\mathbb R$-Banach space, $\Omega\subseteq E_1$ be open, $(M,d)$ be a compact metric space and $f:\Omega\times M\to E_2$ be continuous.
Assume $f(\;\cdot\;,y)$ is Fréchet differentiable for all $y\in M$ and ${\rm D}_1f:\Omega\times M\to\mathfrak L(E_1,E_2)$ is continuous. Are we able to conclude that the differentiability is uniformly with respect to the second argument, i.e. $$\frac1{\left\|h\right\|_{E_1}}\sup_{y\in M}\left\|f(x+h,y)-f(x,y)-{\rm D}_1f(x,y)h\right\|_{E_2}\xrightarrow{h\to0}0\tag1$$ for all $x\in\Omega$.
If necessary, assume that $\Omega$ is relatively compact such that we can infer that $f$ and ${\rm D}_1f$ are uniformly continuous.
Let $\varepsilon>0$. By the plain definition of Fréchet differentiable of $f(\;\cdot\;,y)$ for all $y\in M$, we only obtain that there is a $(\delta_y)_{y\in M}\subseteq(0,\infty)$ with $$\frac{\left\|f(x+h,y)-f(x,y)-{\rm D}_1f(x,y)h\right\|_{E_2}}{\left\|h\right\|_{E_1}}<\varepsilon\tag2$$ for all $h\in B_{\delta_y}(0)$ and $y\in M$.
Example
Assume, for example, that $\Omega=[0,\tau]$ for some $\tau>0$, that $f(\;\cdot\;,x)$ is differentiable for all $x\in M$ and that $\partial_1f$ is (jointly) continuous. Then (since $M$ is compact) $\partial_1f$ is uniformly continuous and hence, given $\varepsilon>0$, there is a $\delta>0$ with $$\left\|\partial_1f(r,x)-\partial_1f(s,y)\right\|_{E_2}<\varepsilon\tag3$$ for all $r,s\in[0,\tau]$ and $x,y\in M$ with $\max(|r-s|,\left\|x-y\right\|_E)<\delta$. Thus, $$\left\|\frac{f(t,x)-f(s,x)}{t-s}-\partial_1f(s,x)\right\|_{E_2}\le\frac1{t-s}\int_s^t\left\|\partial_1f(r,x)-\partial_1f(s,y)\right\|_{E_2}{\rm d}r<\varepsilon\tag4$$ for all $s,t\in[0,\tau]$ with $|s-t|<\delta$ and $x\in M$.
Can we generalize this result?
We have $\frac{1}{\|h\|_E}(f(x+h,y)-f(x,y)-D_1f(x,y)h) = \int_0^1{\left(D_1f(x+th,y)-D_1f(x,y)\right)\frac{h}{\|h\|_E}}$.
Thus, the conclusion follows as soon as $D_1f(x',y)$ converges (in norm) uniformly in $y$, as $x' \rightarrow x$, to $D_1f(x,y)$ (*).
Let $g=D_1f: \Omega \times M \rightarrow E_3$, which is continuous. We'll show that (*) follows from the continuity of $g$.
Indeed, if (*) doesn't hold, as $M$ is compact, we can take sequences $x_n,y_n$ converging to $x,y$ respectively, such that for some $\epsilon>0$, $\|g(x_n,y_n)-g(x,y_n)\| > \epsilon$. But as $g$ is continuous, $g(x_n,y_n) \rightarrow g(x,y)$ and $g(x,y_n) \rightarrow g(x,y)$, a contradiction.