Changing variables in a partial derivative If I have the equation and function $$f_1(x_1,x_2,x_3,...,x_n) = 0,\qquad x_1 = g_1(y_1, y_2, y_3,...,y_m)$$ then what is $\frac {\partial f_1}{\partial x_1}$ in terms of $g_1$ and $y_i$?
 A: Not sure if that is what you need, 
but if you have a function of $n$ variables $f_{1}(x_{1}, x_{2}, x_{3},...,x_{n})$, where 
$x_{1} = g_{1}(y_{1}, y_{2}, y_{3},...,y_{m})$, $x_{2} = g_{2}(y_{1}, y_{2}, y_{3},...,y_{m}),\dots, x_{n} = g_{n}(y_{1}, y_{2}, y_{3},...,y_{m})$  are also some functions of $m$ variables, then you can find partial derivatives of $f_{1}$ with respect to $y_{i},  i=\overline{1,n} $ using the formula:
$$\frac{\partial f_{1}}{\partial y_{i}} = \frac{\partial f_{1}}{\partial x_{1}}\frac{\partial x_{1}}{\partial y_{i}} + \frac{\partial f_{1}}{\partial x_{2}}\frac{\partial x_{2}}{\partial y_{i}} + \ldots + \frac{\partial f_{1}}{\partial x_{n}}\frac{\partial x_{n}}{\partial y_{i}} = \sum_{k=1}^{n} \frac{\partial f_{1}}{\partial x_{k}}\frac{\partial x_{k}}{\partial y_{i}}$$
A: Suppose that you have a function of more than one variable.  Let's take the example of $f$, where:  $$f(x,y)=x^2-2xy+y^2-1$$  This function can take differing values as $x$ and $y$ vary.  And that means we have "interesting" derivatives we can study.
\begin{align*}
\frac{\partial f}{\partial x} & =2x-2y\\
\frac{\partial f}{\partial y} & =-2x+2y
\end{align*}
Sometimes, we will become interested in studying one of $f$'s level sets.  That means the set of all $(x,y)$ such that the output of $f$ is some constant level $k$.  In our example, we might consider $$f(x,y) = 0$$ and what we mean is the set of all $(x,y)$ where $$x^2-2xy+y^2-1=0$$ It doesn't really make sense to speak of $\frac{\partial f}{\partial x}$ anymore.  Our attention is focused on the curve with the equation $x^2-2xy+y^2-1=0$, and on that curve $f$ is identically $0$.  We might consider a small change in $x$, but the constraint will force a small change in $y$ so that together, the corresponding change in $f$ will be $0$.  So if $\frac{\partial f}{\partial x}$ means anything at all, it means the derivative of the zero function: $0$.
Of course nothing is stopping us from going back and working with the partial derivative computed earlier: $\frac{\partial f}{\partial x} =2x-2y$.  And maybe that is what you intend in your question.  But then I ask why include the "$=0$" in your question at all?
Assuming that you would be interested in the partial derivative $2x-2y$ for this example, then the answer is just as Abramodj has said.  To keep this example going, suppose $x=g(s,t)=\sin(s)\cos(t)$.  Then 
\begin{align*}
\frac{\partial f}{\partial x} & = 2x-2y\\
&=2g(s,t)-2y\\
&=2\sin(s)\cos(t)-2y
\end{align*}
May I ask how this question arose?  It looks like a familiar issue that arises in a vector calculus course when you learn that on surfaces $\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}$.  But I might be mistaken.
A: In my opinion it is just
$$ \frac {\partial f_1}{\partial x_1} $$
and the relation you ask is given by substituting $x_1$ in this expression with
$$ g_1(y_1, y_2, y_3,...,y_m) $$
