If I have $z = f(x,y)$ and $x = u(t)$ and $y = v(t)$,
Then, derivative of $z$ w.r.t $t$ can simply be calculated using the following:
$$ \frac{dz}{dt} = \frac{\partial z}{\partial x} * \frac{dx}{dt} + \frac{\partial z}{\partial y} * \frac{dy}{dt}$$
In this case, how do we point out the direction for the above derivative, in case we've a graph plotted for $z = f(x,y)$ in 2-d coordinate system.
As Michael Hoppe stated in the comments, $z=f(x,y)=f(u(t),v(t))$ has only one variable. This means that the direction of the derivative on a 2D graph of $z$ vs $t$ would just be the vector $[1,\frac{dz}{dt}]$.
If we are looking for the direction that $t$ forces $x$ and $y$ to go on the 2D plane of $x$ and $y$, we can see that the curve that is traced out is defined by the parametric $(u(t),v(t))$ and so the direction at a time $t$ would be the vector $[u'(t),v'(t)]$.