746. Find the partial derivatives of the implicit function $z(x, y)$ given by the equations: 1) $x^{2}+y^{2}+z^{2}-z=0$; 2) $a x+b y-c z=$ $=k \cos (a x+b y-c z)$.
Solution. 1) Denoting the left-hand side of the equation by $\varphi(x, y, z)$ and using formulas (A), we get

$$
\frac{\partial z}{\partial x}=-\frac{\varphi_{x}^{\prime}}{\varphi_{z}^{\prime}}=-\frac{2 x}{2 z-1} ; \quad \frac{\partial z}{\partial y}=-\frac{\varphi_{y}^{\prime}}{\varphi_{z}^{\prime}}=-\frac{2 y}{2 z-1}
$$

2) Transforming the equation to the form $a x+b y-c z-k \cos (a x+$ $+b y-c z)=0$ and denoting its left-hand side by $F(x, y, z)$, using formulas (A) we find

$$
\begin{aligned}
& \frac{\partial z}{\partial x}=-\frac{F_{x}^{\prime}}{F_{z}^{\prime}}=-\frac{a+a k \sin (a x+b y-c z)}{-c-c k \sin (a x+b y-c z)}=\frac{a}{c} \\
& \frac{\partial z}{\partial y}=-\frac{F_{y}^{\prime}}{F_{z}^{\prime}}=-\frac{b+b k \sin (a x+b y-c z)}{-c-c k \sin (a x+b y-c z)}=\frac{b}{c}
\end{aligned}
$$

Find the derivatives of implicit functions: