1. **State the problem:** We are given the function $$f(x,y) = \frac{y^2 + 2}{2 e^{3x} - 1}$$ and need to find the partial derivatives $$f_x$$, $$f_y$$, and the mixed partial derivative $$f_{yx}$$. 2. **Recall the formulas and rules:** - The partial derivative with respect to $$x$$ treats $$y$$ as a constant. - The partial derivative with respect to $$y$$ treats $$x$$ as a constant. - Use the quotient rule or rewrite the function to apply the chain and product rules. 3. **Find $$f_x$$:** Rewrite $$f(x,y) = (y^2 + 2)(2 e^{3x} - 1)^{-1}$$. Apply the product and chain rules: $$ f_x = (y^2 + 2) \cdot \frac{d}{dx} \left( (2 e^{3x} - 1)^{-1} \right) $$ Derivative inside: $$ \frac{d}{dx} (2 e^{3x} - 1)^{-1} = -1 \cdot (2 e^{3x} - 1)^{-2} \cdot \frac{d}{dx} (2 e^{3x} - 1) $$ Calculate $$\frac{d}{dx} (2 e^{3x} - 1) = 6 e^{3x}$$. So, $$ f_x = (y^2 + 2)(-1)(2 e^{3x} - 1)^{-2} (6 e^{3x}) = -\frac{6 e^{3x} (y^2 + 2)}{(2 e^{3x} - 1)^2} $$ 4. **Find $$f_y$$:** Since $$x$$ is constant, $$ f_y = \frac{\partial}{\partial y} \left( \frac{y^2 + 2}{2 e^{3x} - 1} \right) = \frac{2y}{2 e^{3x} - 1} $$ 5. **Find $$f_{yx}$$:** Take the partial derivative with respect to $$x$$ of $$f_y$$: $$ f_y = \frac{2y}{2 e^{3x} - 1} = 2y (2 e^{3x} - 1)^{-1} $$ Apply chain rule: $$ f_{yx} = 2y \cdot (-1)(2 e^{3x} - 1)^{-2} \cdot 6 e^{3x} = -\frac{12 y e^{3x}}{(2 e^{3x} - 1)^2} $$ **Final answers:** $$ f_x = -\frac{6 e^{3x} (y^2 + 2)}{(2 e^{3x} - 1)^2}, \quad f_y = \frac{2y}{2 e^{3x} - 1}, \quad f_{yx} = -\frac{12 y e^{3x}}{(2 e^{3x} - 1)^2} $$