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We are tasked with finding the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) given the function:

\[
z = x^2 + \tan^{-1}\left(\frac{y}{x}\right)
\]

### 1. Finding \(\frac{\partial z}{\partial x}\):

The function \(z(x, y)\) consists of two terms:

\[
z = x^2 + \tan^{-1}\left(\frac{y}{x}\right)
\]

#### First term: \(x^2\)
The derivative of \(x^2\) with respect to \(x\) is:

\[
\frac{\partial}{\partial x}(x^2) = 2x
\]

#### Second term: \(\tan^{-1}\left(\frac{y}{x}\right)\)
To differentiate this term, we apply the chain rule. Let \( u = \frac{y}{x} \). Then:

\[
\frac{\partial}{\partial x} \left( \tan^{-1}(u) \right) = \frac{1}{1 + u^2} \cdot \frac{\partial u}{\partial x}
\]

Now, we need \(\frac{\partial u}{\partial x}\), where \(u = \frac{y}{x}\):

\[
\frac{\partial}{\partial x} \left( \frac{y}{x} \right) = -\frac{y}{x^2}
\]

Thus:

\[
\frac{\partial}{\partial x} \left( \tan^{-1}\left(\frac{y}{x}\right) \right) = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right)
\]

Simplifying this expression:

\[
\frac{\partial}{\partial x} \left( \tan^{-1}\left(\frac{y}{x}\right) \right) = \frac{-y}{x^2 + y^2}
\]

#### Combine both terms:
\[
\frac{\partial z}{\partial x} = 2x - \frac{y}{x^2 + y^2}
\]

### 2. Finding \(\frac{\partial z}{\partial y}\):

#### First term: \(x^2\)
Since \(x^2\) is independent of \(y\), its derivative with respect to \(y\) is:

\[
\frac{\partial}{\partial y}(x^2) = 0
\]

#### Second term: \(\tan^{-1}\left(\frac{y}{x}\right)\)
Again, let \(u = \frac{y}{x}\). The derivative with respect to \(y\) is:

\[
\frac{\partial}{\partial y} \left( \tan^{-1}(u) \right) = \frac{1}{1 + u^2} \cdot \frac{\partial u}{\partial y}
\]

Now, \(\frac{\partial u}{\partial y} = \frac{1}{x}\). Thus:

\[
\frac{\partial}{\partial y} \left( \tan^{-1}\left(\frac{y}{x}\right) \right) = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{1}{x}
\]

Simplifying:

\[
\frac{\partial}{\partial y} \left( \tan^{-1}\left(\frac{y}{x}\right) \right) = \frac{1}{x^2 + y^2}
\]

#### Combine both terms:
\[
\frac{\partial z}{\partial y} = \frac{1}{x^2 + y^2}
\]

### Final Answers:
\[
\frac{\partial z}{\partial x} = 2x - \frac{y}{x^2 + y^2}
\]
\[
\frac{\partial z}{\partial y} = \frac{1}{x^2 + y^2}
\]

Would you like a more detailed explanation of any of these steps?

Here are five related questions to further your understanding:
1. How does the chain rule work when differentiating functions involving both \(x\) and \(y\)?
2. What is the geometric interpretation of partial derivatives in a multivariable function?
3. How would the derivatives change if \(z\) included a term like \(\sin(xy)\)?
4. What role does the arctangent function play in calculating these derivatives?
5. How do partial derivatives help in finding the gradient of a function?

**Tip:** When differentiating a function with multiple variables, treat all other variables as constants when focusing on one variable at a time.

Find ∂z/∂x and ∂z/∂y if z = x^2 + tan^(-1)(y/x).

This problem demonstrates how to calculate the partial derivatives of z = x^2 + tan^(-1)(y/x) with respect to x and y, using the chain rule and inverse trigonometric differentiation. The solution provides step-by-step guidance for understanding these key concepts in multivariable calculus.

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