`sin(xy) + (x)/(y) = x^(2) - y`

Explanation

To solve the equation \( \sin(xy) + \frac{x}{y} = x^2 - y \) and find the derivative \( \frac{dy}{dx} \), we will differentiate both sides of the equation with respect to \( x \).

### Step-by-Step Solution:

1. **Differentiate both sides**:
We start by differentiating the equation:
\[ \frac{d}{dx} \left( \sin(xy) + \frac{x}{y} \right) = \frac{d}{dx} \left( x^2 - y \right) \]

2. **Apply the chain rule and product rule**:
For the left side:
- The derivative of \( \sin(xy) \) using the chain rule is:
\[ \cos(xy) \cdot \frac{d}{dx}(xy) = \cos(xy) \left( y + x \frac{dy}{dx} \right) \]
- The derivative of \( \frac{x}{y} \) using the quotient rule is:
\[ \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} = \frac{y - x \frac{dy}{dx}}{y^2} \]

So, the left side becomes:
\[ \cos(xy) \left( y + x \frac{dy}{dx} \right) + \frac{y - x \frac{dy}{dx}}{y^2} \]

For the right side:
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( -y \) is \( -\frac{dy}{dx} \).

Thus, the right side becomes:
\[ 2x - \frac{dy}{dx} \]

3. **Set the derivatives equal**:
Now we have:
\[ \cos(xy) \left( y + x \frac{dy}{dx} \right) + \frac{y - x \frac{dy}{dx}}{y^2} = 2x - \frac{dy}{dx} \]

4. **Rearranging the equation**:
Combine all terms involving \( \frac{dy}{dx} \) on one side:
\[ \cos(xy) y + \frac{y}{y^2} - 2x = -\frac{dy}{dx} - \cos(xy) x \frac{dy}{dx} + \frac{x}{y^2} \frac{dy}{dx} \]

5. **Factor out \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} \left( -\cos(xy) x + \frac{x}{y^2} \right) = 2x - \cos(xy) y - \frac{y}{y^2} \]

6. **Solve for \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} = \frac{2x - \cos(xy) y - \frac{1}{y}}{-\cos(xy) x + \frac{x}{y^2}} \]

### Final Result:
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{2x - \cos(xy) y - \frac{1}{y}}{-\cos(xy) x + \frac{x}{y^2}} \]