find $\frac{\partial f(u(x(t),y(t)),v(x(t),y(t)))}{\partial t} $ how to find 
$$\frac{d z}{d t} $$
where is $z=f(u(x(t),y(t)),v(x(t),y(t)))$
and if some one can give me some advice which make me deal more easy with this subject
 A: My advice for these types of questions is that you learn the multivariable chain rule using the derivative matrix.  This way, as long as you follow the matrix multiplication, you'll get the right answer.
In this case, we have three nested functions.  Define $g(x)=\langle u(x,y),v(x,y) \rangle$ and $h(t) =\langle x(t),y(t)\rangle$, and we can write the function as
$$
z=f(u(x(t),y(t)),v(x(t),y(t))) = f\circ g\circ h (t)
$$
Now, the matrix derivative chain-rule gives us the following:
$$
\begin{align}
D f\circ g\circ h (t) &= \frac{\partial f}{\partial t}\\
&= D (f\circ g\circ h)\\
&= (Df)(g\circ h(t))D (g\circ h) \\
& =(Df)(g\circ h(t))(D g)(h(t)) (Dh)\\
& = \nabla f(g\circ h(t)) \cdot 
\left(\frac{\partial u,v}{\partial x,y}
\begin{bmatrix}
x'(t)\\
y'(t)
\end{bmatrix}
\right)
\end{align}
$$
Where $\nabla$ is the gradient, $\cdot$ is the dot product, and $\frac{\partial u,v}{\partial x,y}$ is the Jacobian matrix, which is matrix multiplied by the matrix on its right.
I hope that helps.
When you expand that product, it looks like this:
$$
\frac{d f}{d t} =\frac{\partial f}{\partial u}
\left(
\frac{\partial u}{\partial x}\frac{dx}{dt} + 
\frac{\partial u}{\partial y}\frac{dy}{dt}
\right) +
\frac{\partial f}{\partial v}
\left(
\frac{\partial v}{\partial x}\frac{dx}{dt} + 
\frac{\partial v}{\partial y}\frac{dy}{dt}
\right)
$$
A: Since $z$ depends only on $t$, we have
$\frac{\partial z}{\partial t} = \frac{dz}{dt}$,
and from there we just use the chain rule a few times.  First we note that
$\frac{dz}{dt} = \frac{\partial f}{\partial u} \frac{du}{dt} + \frac{\partial f}{\partial v} \frac{dv}{dt}$,
then obtain $\frac{du}{dt}$ as
$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}$,
with a similar expression holding for $\frac{dv}{dt}$:
$\frac{dv}{dt} = \frac{\partial v}{\partial x} \frac{dx}{dt} + \frac{\partial v}{\partial y}\frac{dy}{dt}$.
From here on out, it's just algebra, recombining terms:
$\frac{\partial z}{\partial t} = \frac{dz}{dt} = \frac{\partial f}{\partial u}(\frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}) + \frac{\partial f}{\partial v} (\frac{\partial v}{\partial x} \frac{dx}{dt} + \frac{\partial v}{\partial y}\frac{dy}{dt})$
$= (\frac{\partial f}{\partial u} \frac{\partial u}{\partial x} +  \frac{\partial f}{\partial v} \frac{\partial v}{\partial x})\frac{dx}{dt} + (\frac{\partial f}{\partial u} \frac{\partial u}{\partial y} +  \frac{\partial f}{\partial v} \frac{\partial v}{\partial y})\frac{dy}{dt}$.
You can't say much more without knowing the specific forms of $f(u, v)$, $u(x, y)$, $v(x, y)$, $x(t)$ and $y(t)$.
Hope this clarifies.  Cheers.
