If `sin(x y)+y/x=x^2-y^2` , find `(dy)/(dx)` .

Explanation

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

### Step-by-Step Solution:

1. **Differentiate both sides**:
We start by differentiating the equation \( \sin(xy) + \frac{y}{x} = x^2 - y^2 \) with respect to \(x\).

\[ \frac{d}{dx} \left( \sin(xy) \right) + \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{d}{dx} \left( x^2 \right) - \frac{d}{dx} \left( y^2 \right) \]

2. **Apply the chain rule and product rule**:
- For \( \sin(xy) \), we use the chain rule:
\[ \frac{d}{dx} \left( \sin(xy) \right) = \cos(xy) \cdot \frac{d}{dx}(xy) = \cos(xy) \left( y + x \frac{dy}{dx} \right) \]
- For \( \frac{y}{x} \), we use the quotient rule:
\[ \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{x \frac{dy}{dx} - y}{x^2} \]
- The derivatives of \(x^2\) and \(y^2\) are:
\[ \frac{d}{dx}(x^2) = 2x \quad \text{and} \quad \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]

3. **Combine the derivatives**:
Putting it all together, we have:
\[ \cos(xy) \left( y + x \frac{dy}{dx} \right) + \frac{x \frac{dy}{dx} - y}{x^2} = 2x - 2y \frac{dy}{dx} \]

4. **Clear the fractions**:
Multiply through by \(x^2\) to eliminate the fraction:
\[ x^2 \cos(xy) \left( y + x \frac{dy}{dx} \right) + x \frac{dy}{dx} - y = 2x^3 - 2xy \frac{dy}{dx} \]

5. **Expand and collect terms**:
Expanding gives:
\[ x^2 y \cos(xy) + x^3 \cos(xy) \frac{dy}{dx} + x \frac{dy}{dx} - y = 2x^3 - 2xy \frac{dy}{dx} \]
Rearranging terms gives:
\[ (x^3 \cos(xy) + x + 2xy) \frac{dy}{dx} = 2x^3 + y - x^2 y \cos(xy) \]

6. **Solve for \(\frac{dy}{dx}\)**:
Finally, we isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2x^3 + y - x^2 y \cos(xy)}{x^3 \cos(xy) + x + 2xy} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{2x^3 + y - x^2 y \cos(xy)}{x^3 \cos(xy) + x + 2xy} \]