745. Find the derivative of the implicit function $y$, given by the equation: 1) $x^{2}+y^{2}+2 x-6 y+2=0$; 2) $x^{y}=y^{x}$, and calculate its value at $x=1$.
Solution. 1) Denoting the left-hand side of the given equation by $f(x, y)$, we find the partial derivatives $f_{x}^{\prime}=2 x+2, f_{y}^{\prime}=2 y-6$ and, substituting them into formula (B), we get $y^{\prime}=\frac{x+1}{3-y}$.

Next, substituting $x=1$ into the original equation, we find two corresponding values of the function $y_{1}=1$ and $y_{2}=5$. Therefore, at $x=1$, the derivative has two values: $y_{1}^{\prime}(1)=1, y_{2}^{\prime}(1)=-1$.

2) Transforming the given equation to the form $x^{y}-y^{x}=0$, according to formula (B), we get

$$
\frac{d!y}{d x}=-\frac{\left(x^{y}-y^{x}\right)_{x}^{\prime}}{\left(x^{y}-y^{x}\right)_{!}^{\prime}}=\frac{y^{x} \ln y-y x^{y-1}}{x^{y} \ln x-x y^{x-1}}
$$

When $x=1$, from the given equation we determine $y=-1$. The sought value is $y^{\prime}(1)=1$.