(5)z=lntan(x)/(y); (6)z=(1+xy)^y; (7)u=x^(y)/(2); (8)u=arctan(x-y)^2.

(5) $ z = \ln \tan \frac{x}{y} $ \[ \boxed{ \begin{aligned} &\frac{\partial z}{\partial x} = \frac{2}{y} \csc \frac{2x}{y}, \\ &\frac{\partial z}{\partial y} = -\frac{2x}{y^2} \csc \frac{2x}{y}. \end{aligned} } \] (6) $ z = (1+xy)^y $ \[ \boxed{ \begin{aligned} &\frac{\partial z}{\partial x} = y^2 (1+xy)^{y-1}, \\ &\frac{\partial z}{\partial y} = (1+xy)^y \left[ \ln (1+xy) + \frac{xy}{1+xy} \right]. \end{aligned} } \] (7) $ u = x^{\frac{y}{z}} $ \[ \boxed{ \begin{aligned} &\frac{\partial u}{\partial x} = \frac{y}{z} x^{\frac{y}{z} - 1}, \\ &\frac{\partial u}{\partial y} = x^{\frac{y}{z}} \cdot \frac{\ln x}{z}, \\ &\frac{\partial u}{\partial z} = -\frac{y \ln x}{z^2} x^{\frac{y}{z}}. \end{aligned} } \] (8) $ u = \arctan (x-y)^x $ \[ \boxed{ \begin{aligned} &\frac{\partial u}{\partial x} = \frac{(x-y)^x \left[ \ln (x-y) + \frac{x}{x-y} \right]}{1+(x-y)^{2x}}, \\ &\frac{\partial u}{\partial y} = -\frac{x (x-y)^{x-1}}{1+(x-y)^{2x}}. \end{aligned} } \]