【解】由复合函数求导法则，$z=u^2+\cos(xy+u)$，其中 $u=x\ln y$。

先计算 $\dfrac{\partial z}{\partial x}$：
$$
\frac{\partial z}{\partial x} = 2u \cdot \frac{\partial u}{\partial x} - \sin(xy+u) \cdot \left(y + \frac{\partial u}{\partial x}\right),
$$
其中 $\dfrac{\partial u}{\partial x} = \ln y$，代入得
$$
\frac{\partial z}{\partial x} = 2x\ln y \cdot \ln y - \sin(xy+x\ln y) \cdot (y + \ln y) = 2x(\ln y)^2 - (y + \ln y)\sin(xy + x\ln y).
$$

再计算 $\dfrac{\partial z}{\partial y}$：
$$
\frac{\partial z}{\partial y} = 2u \cdot \frac{\partial u}{\partial y} - \sin(xy+u) \cdot \left(x + \frac{\partial u}{\partial y}\right),
$$
其中 $\dfrac{\partial u}{\partial y} = \dfrac{x}{y}$，代入得
$$
\frac{\partial z}{\partial y} = 2x\ln y \cdot \frac{x}{y} - \sin(xy+x\ln y) \cdot \left(x + \frac{x}{y}\right)
= \frac{2x^2\ln y}{y} - x\left(1 + \frac{1}{y}\right)\sin(xy + x\ln y).
$$
提取公因子 $\dfrac{x}{y}$：
$$
\frac{\partial z}{\partial y} = \frac{x}{y}\left[2x\ln y - (y+1)\sin(xy + x\ln y)\right].
$$
答案为 $\dfrac{\partial z}{\partial x}=2x(\ln y)^2-(y+\ln y)\sin(xy+x\ln y)$，$\dfrac{\partial z}{\partial y}=\dfrac{x}{y}\left[2x\ln y-(y+1)\sin(xy+x\ln y)\right]$。