Let $a,b : I \to \mathbb{R}$ be differentiable functions, and $m: I \to \mathbb{R}^2$ be a function such that $m(t)=(a(t),b(t))$.
Further, let there be a differentiable function $f:\mathbb{R}^2\to\mathbb{R}$, and let $z:I\to\mathbb{R}$, $z(t)=f(a(t),b(t))$.
We wish to find $\frac{dz}{dt}$.
We know that $\frac{dz}{dt}=D(z)(t)$. (Total derivative)
So $$D(z)(t)=D(f\circ m)(t)=D(f)(m(t)).D(m)(t)$$ (Chain rule) $$D(f)(m(t))=[\frac{\partial f}{\partial x}(m(t)) \space\space\space\frac{\partial f}{\partial y}(m(t))]$$ By this I mean, partial derivative of $f$ w.r.t $x$ at $m(t)$. $$ D(m)(t)= \left[ {\begin{array}{c} \frac{da}{dt} \\ \frac{db}{dt} \\ \end{array} } \right] $$. Multiplying both gives, $$\frac{dz}{dt}=\frac{\partial f}{\partial x}.\frac{da}{dt}+\frac{\partial f}{\partial y}.\frac{db}{dt}$$ Now how to express this in terms of $z$ only not $f$?
According to the book I have, the answer is $$\frac{dz}{dt}=\frac{\partial z}{\partial a}.\frac{da}{dt}+\frac{\partial z}{\partial b}.\frac{db}{dt}$$ Also, what is the meaning of $\frac{\partial z}{\partial a}$ according to the definition?