To verify that $\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y}$, where $z = \sin^{-1} \left( \frac{x}{y} \right)$, let's go through the steps to compute each of these mixed partial derivatives.

Step 1: Define $z = \sin^{-1} \left( \frac{x}{y} \right)$

Since $z = \sin^{-1} \left( \frac{x}{y} \right)$, we can rewrite it in terms of partial derivatives with respect to $x$ and $y$.

Step 2: Compute $\frac{\partial z}{\partial x}$

Using the chain rule for derivatives, we get: $\frac{\partial z}{\partial x} = \frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \frac{\partial}{\partial x} \left( \frac{x}{y} \right)$ Since $\frac{x}{y}$ is a function of $x$ and $y$, we differentiate $\frac{x}{y}$ with respect to $x$, treating $y$ as a constant: $\frac{\partial z}{\partial x} = \frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \frac{1}{y} = \frac{1}{y \sqrt{1 - \frac{x^2}{y^2}}}$ Simplifying further, this becomes: $\frac{\partial z}{\partial x} = \frac{y}{y^2 - x^2}$

Step 3: Compute $\frac{\partial^2 z}{\partial y \partial x}$

Now, take the derivative of $\frac{\partial z}{\partial x}$ with respect to (