To find the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\), we will use the chain rule for multivariable functions. The function \(f(x, y) = g(xy, \frac{x}{y})\) is composed of the function \(g(u, v)\) where \(u = xy\) and \(v = \frac{x}{y}\). We will first express the partial derivatives of \(f\) in terms of the partial derivatives of \(g\) with respect to \(u\) and \(v\), and then compute these derivatives.
Dada la función \( f(x, y) = g(xy, \frac{x}{y}) \), donde \( g: \mathbb{R}^{2} \rightarrow \mathbb{R} \) es de clase \( C^{2} \), definimos las variables intermedias:
\[
u = xy \quad \text{y} \quad v = \frac{x}{y}
\]
Utilizando la regla de la cadena, la derivada parcial de \( f \) con respecto a \( x \) se expresa como:
\[
\frac{\partial f}{\partial x} = \frac{\partial g}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial g}{\partial v} \cdot \frac{\partial v}{\partial x}
\]
Calculando las derivadas parciales de \( u \) y \( v \):
\[
\frac{\partial u}{\partial x} = y \quad \text{y} \quad \frac{\partial v}{\partial x} = \frac{1}{y}
\]
Por lo tanto, la derivada parcial se convierte en:
\[
\frac{\partial f}{\partial x} = y \cdot \frac{\partial g}{\partial u} + \frac{1}{y} \cdot \frac{\partial g}{\partial v}
\]
De manera similar, la derivada parcial de \( f \) con respecto a \( y \) es:
\[
\frac{\partial f}{\partial y} = \frac{\partial g}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial g}{\partial v} \cdot \frac{\partial v}{\partial y}
\]
Calculando las derivadas parciales de \( u \) y \( v \):
\[
\frac{\partial u}{\partial y} = x \quad \text{y} \quad \frac{\partial v}{\partial y} = -\frac{x}{y^2}
\]
Por lo tanto, la derivada parcial se convierte en:
\[
\frac{\partial f}{\partial y} = x \cdot \frac{\partial g}{\partial u} - \frac{x}{y^2} \cdot \frac{\partial g}{\partial v}
\]
Las derivadas parciales son:
\[
\frac{\partial f}{\partial x} = y \cdot \frac{\partial g}{\partial u} + \frac{1}{y} \cdot \frac{\partial g}{\partial v}
\]
\[
\frac{\partial f}{\partial y} = x \cdot \frac{\partial g}{\partial u} - \frac{x}{y^2} \cdot \frac{\partial g}{\partial v}
\]
Por lo tanto, la respuesta final es:
\[
\boxed{\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)}
\]
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