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好的，我们来求解这些函数的全微分。

### (1) \( z = xy + \frac{x}{y} \)

首先计算 \( z \) 对 \( x \) 和 \( y \) 的偏导数：
\[ \frac{\partial z}{\partial x} = y + \frac{1}{y} \]
\[ \frac{\partial z}{\partial y} = x - \frac{x}{y^2} \]

全微分为：
\[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \]
\[ dz = \left( y + \frac{1}{y} \right) dx + \left( x - \frac{x}{y^2} \right) dy \]

### (2) \( z = e^{\frac{y}{x}} \)

计算 \( z \) 对 \( x \) 和 \( y \) 的偏导数：
\[ \frac{\partial z}{\partial x} = e^{\frac{y}{x}} \cdot \left( -\frac{y}{x^2} \right) = -\frac{y}{x^2} e^{\frac{y}{x}} \]
\[ \frac{\partial z}{\partial y} = e^{\frac{y}{x}} \cdot \frac{1}{x} = \frac{1}{x} e^{\frac{y}{x}} \]

全微分为：
\[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \]
\[ dz = \left( -\frac{y}{x^2} e^{\frac{y}{x}} \right) dx + \left( \frac{1}{x} e^{\frac{y}{x}} \right) dy \]

### (3) \( z = \frac{y}{\sqrt{x^2 + y^2}} \)

计算 \( z \) 对 \( x \) 和 \( y \) 的偏导数：
\[ \frac{\partial z}{\partial x} = \frac{y \cdot \left( -\frac{x}{\sqrt{x^2 + y^2}} \right)}{x^2 + y^2} = -\frac{xy}{(x^2 + y^2)^{3/2}} \]
\[ \frac{\partial z}{\partial y} = \frac{\sqrt{x^2 + y^2} - y \cdot \frac{y}{\sqrt{x^2 + y^2}}}{x^2 + y^2} = \frac{x^2}{(x^2 + y^2)^{3/2}} \]

全微分为：
\[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \]
\[ dz = -\frac{xy}{(x^2 + y^2)^{3/2}} dx + \frac{x^2}{(x^2 + y^2)^{3/2}} dy \]

请问您需要详细的推导过程吗？还有其他问题需要解答吗？

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小提示：在求解全微分时，熟练掌握偏导数的计算方法非常关键。

Learn how to solve total differentials for multivariable functions with detailed step-by-step solutions. This problem covers key concepts in partial differentiation and total differentials. Suitable for undergraduate students.

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