If the straight line `y=x` meets `y=f(x)` at P, where `f(x)` is a solution of the differential equation `(dy)/(dx)=(x^(2)+xy)/(x^(2)+y^(2))` such that `f(1)=3`, then the value of `f'(x)` at the point P is

Options

A. `(3)/(5)`
B. `(5)/(3)`
C. 2
D. 1

Answer

To solve the problem, we need to find the value of \( f'(x) \) at the point \( P \) where the line \( y = x \) intersects the function \( y = f(x) \), given that \( f(x) \) is a solution to the differential equation

\[ \frac{dy}{dx} = \frac{x^2 + xy}{x^2 + y^2} \]

and that \( f(1) = 3 \).

### Step-by-Step Solution

1. **Identify the point of intersection \( P \)**:
Since the line \( y = x \) intersects \( y = f(x) \) at point \( P \), at this point, we have:
\[ f(x) = x \]

2. **Substituting \( y = f(x) \) into the differential equation**:
We have the differential equation:
\[ \frac{dy}{dx} = \frac{x^2 + xy}{x^2 + y^2} \]
Substituting \( y = f(x) \), we can rewrite it as:
\[ \frac{dy}{dx} = \frac{x^2 + x f(x)}{x^2 + f(x)^2} \]

3. **Finding \( f'(x) \) at point \( P \)**:
We need to evaluate \( f'(x) \) at the point where \( f(x) = x \). This means we will substitute \( f(x) = x \) into the differential equation:
\[ \frac{dy}{dx} = \frac{x^2 + x^2}{x^2 + x^2} = \frac{2x^2}{2x^2} = 1 \]
Thus, at the point \( P \), we find:
\[ f'(x) = 1 \]

4. **Evaluate at the specific point \( P \)**:
We know that \( f(1) = 3 \), which means the point \( P \) is actually \( (1, 3) \). However, since we are looking for \( f'(x) \) at the intersection point where \( y = x \), we have already established that \( f'(x) = 1 \) at that intersection.

### Final Answer
The value of \( f'(x) \) at point \( P \) is:
\[ \boxed{1} \]