Answer
$\displaystyle \frac{\partial z}{\partial x}=-\frac{x}{3z},\qquad \frac{\partial z}{\partial y}=-\frac{2y}{3z}$.
Work Step by Step
$x^{2}+2y^{2}+3z^{2}=1 \displaystyle \qquad/\frac{\partial}{\partial x}$, (y is constant, z is a function of x)
$\displaystyle \frac{\partial}{\partial x}(x^{2}+2y^{2}+3z^{2})=\frac{\partial}{\partial x}(1)$
$2x+0+6z \displaystyle \frac{\partial z}{\partial x}=0$
$6z\displaystyle \frac{\partial z}{\partial x}=-2x$
$\displaystyle \frac{\partial z}{\partial x}=\frac{-2x}{6z}$
$\displaystyle \frac{\partial z}{\partial x}=-\frac{x}{3z}$
$x^{2}+2y^{2}+3z^{2}=1 \displaystyle \qquad/\frac{\partial}{\partial y}$, (x is constant, z is a function of y)
$\displaystyle \frac{\partial}{\partial y}(x^{2}+2y^{2}+3z^{2})=\frac{\partial}{\partial y}(1)$
$0+4y+6z\displaystyle \frac{\partial z}{\partial y}=0$
$6z \displaystyle \frac{\partial z}{\partial y}=-4y$
$ \displaystyle \frac{\partial z}{\partial y}=\frac{-4y}{6z}$
$ \displaystyle \frac{\partial z}{\partial y}=-\frac{2y}{3z}$.
