I have the function $u(x,t)= u(x(y,z), t(y,z))$ where $x(y,z)$ and $t(y, z)$ are functions of $y$ and $z$.
I don't understand the following equality:
$\dfrac{\partial u}{\partial y} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial y}+ \dfrac{\partial u}{\partial t}\dfrac{\partial t}{\partial y}$
This is just the multi-dimensional chain rule.
First, just look at $u$ as a function of $x$ and $t$. If we make a small change $dx$ in the value of $x$ while holding $t$ constant, then we get a change $$(x + dx, t) - u(x, t) \approx \frac{\partial u}{\partial x}dx$$ in the value of $u$. Similarly if we change $t$ while holding $x$ constant we get a change $$u(x, t + dt) - u(x, t) \approx \frac{\partial u}{\partial t}dt$$ So what happens if we change both $x$ and $t$? Since $u$ is differentiable, you can show (I won't, because I am trying to explain the idea, not the details) that $$u(x + dx, t) - u(x, t) \approx u(x + dx, t + dt) - u(x, t + dt)$$ That is, sliding the first equation over from using $t$ as the fixed value of the 2nd variable to using $t + dt$ as that value makes only a negligible difference in the change in value of $u$. Thus $$\begin{align}du &= u(x + dx, t+dt) - u(x, t)\\ &= u(x + dx, t+dt) - u(x, t+dt) + u(x, t+dt) - u(x,t)\\ &\approx u(x + dx, t) - u(x, t)+ u(x, t+dt) - u(x,t)\\ &\approx \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial t}dt\end{align}$$ The infinitesimal version is the differential equation $$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial t}dt$$
Now what happens if we make $x$ and $t$ both be functions of $y$? Then a small change $dy$ in the value of $y$ results in a small change $dx$ in the value of $x$ and $dt$ in the value of $t$, since both vary with $y$. But $$dx = \frac {dx}{dy}dy, \quad dt = \frac{dt}{dy}dy$$ So we get $$\begin{align}du &= \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial t}dt\\[3pt]&=\frac{\partial u}{\partial x}\frac {dx}{dy}dy + \frac{\partial u}{\partial t}\frac{dt}{dy}dy\end{align}$$ Dividing by $dy$ and taking the limit, $$\frac{du}{dy}=\frac{\partial u}{\partial x}\frac {dx}{dy} + \frac{\partial u}{\partial t}\frac{dt}{dy}$$
Your equation is just the same, except that you have introduced a second variable $z$, so the derivatives become partial derivatives holding $z$ constant.