Show that if $f(t,x,\varepsilon)$ is differentiable with respect to $\varepsilon$ and $f(t,0,\varepsilon)=0$, then $f$ is lipschitz in $\varepsilon$

Question

Suppose

1. $f(t,x,\varepsilon)$ is bounded and has bounded partial derivatives up to the second order with respect to its arguments, $(t,x, \varepsilon)$, in a neighborhood of $x=0$.
2. $f(t,0,\varepsilon)=0$ for all $t>0$ and $\varepsilon$, where $\varepsilon>0$ is a small parameter.

Then I want to show that $$\|f(t,x,\varepsilon)-f(t,x,0)\| \leq L \varepsilon \|x\|$$ where $L>0$ is a constant. In other words, $f$ is Lipschitz with respect to $\varepsilon$, linearly in $x$.

We may not need all of the assumptions stated above. But I think we certainly need $f(t,0,\varepsilon)=0$ and $f$ to be differentiable with respect to $\varepsilon$.