Interchange limit of one variable with partial derivative of another variable

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Consider a function $f(t,y)$ where $t,y \in \mathbb{R}$.

It is given that $ f(t,y)$ is continuous but $\dfrac{\partial f(t,y)}{\partial t}$ suffers a jump at at $t= x$. (The function $f$ could describe the temperature field in a composite region where different metal strips are glued together at $t=x$ )

How can I then show that $\dfrac{\partial f(t,y)}{\partial y}$ is continuous at $t=x$ ?

My attempt:

We need to show $$\lim_{t \rightarrow x^{-}} \dfrac{\partial f(t,y)}{\partial y} = \lim_{t \rightarrow x^{+}} \dfrac{\partial f(t,y)}{\partial y}$$ This is equivalent to showing $$\lim_{t \rightarrow x^{-}} \lim_{h\rightarrow 0}\dfrac{ f(t,y+h) - f(t, y)}{h} = \lim_{t \rightarrow x^{+}} \lim_{h\rightarrow 0}\dfrac{ f(t,y+h) - f(t, y)}{h}$$

Q) Now is there some hypothesis on $f(t,y)$ or $\dfrac{\partial f(t,y)}{\partial y}$ that will allow me to interchange the order of limits ?

If so, then it is enough to show $$\lim_{h\rightarrow 0} \lim_{t \rightarrow x^{-}} \dfrac{ f(t,y+h) - f(t, y)}{h} = \lim_{h\rightarrow 0} \lim_{t \rightarrow x^{+}} \dfrac{ f(t,y+h) - f(t, y)}{h}$$

By continuity of $f(t,.)$ at $t=x$,

$$\lim_{t \rightarrow x^{-}} \dfrac{ f(t,y+h) - f(t, y)}{h} = \lim_{t \rightarrow x^{+}} \dfrac{ f(t,y+h) - f(t, y)}{h} = \dfrac{ f(x,y+h) - f(x, y)}{h}$$ which proves what I want.

So my question is exactly what hypothesis/theorem is necessary to permit the interchange of limits here ?

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You have not clearly stated what the assumptions with respect to ${\partial\over\partial y}$ are. The function $f(t,y):=|y|$ satisfies your assumptions, but $f_y$ is not continuous, resp., undefined at $y=0$.

One possible setup could be the following: $$f_t(t,y)=\cases{g_>(t,y)\quad&$(t>0)$ \cr g_<(t,y)\quad&$(t<0)$\cr}$$ with $g_>$, $g_<\in C^1$ in a full neighborhood of $(0,0)$; furthermore we assume that $\phi(y):=f(0,y)$ is $C^1$ in a neighborhood of $0$. Then one has $$f(t,y)=\cases{f(0,y)+\int_0^t g_>(s,y)\>ds\quad&$(t\geq0)$ \cr f(0,y)+\int_0^t g_<(s,y)\>ds\quad&$(t\leq0)$\cr}\ ,$$ and it becomes obvious that $f_y\in C^1$ in a neighborhood of $(0,0)$.