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To solve for \( f(x, y) \), we need to use the given partial differential equation and the additional condition. The problem states:

1. \( f_x(x+y) = x^2 + y^2 + 2xy - 2x \)
2. \( f(y, y) = -y^3 \)

We are to find \( f(x, y) \).

### Step-by-Step Solution:

1. **Express the given PDE in terms of a single variable transformation:**

   Let \( u = x + y \). Therefore, \( \frac{\partial f}{\partial x} = f_x \) can be written as \( f_x(u) \).

2. **Substitute \( u \) in the given PDE:**

   \( f_x(u) = x^2 + y^2 + 2xy - 2x \)

3. **Express \( x \) and \( y \) in terms of \( u \) and one of the variables:**

   Since \( u = x + y \), we can express \( x = u - y \).

4. **Rewrite the PDE:**

   \[
   f_x(u) = (u-y)^2 + y^2 + 2(u-y)y - 2(u-y)
   \]

   Simplify this expression:

   \[
   f_x(u) = (u^2 - 2uy + y^2) + y^2 + 2uy - 2y^2 - 2u + 2y
   \]

   \[
   f_x(u) = u^2 - 2u + 2y
   \]

5. **Integrate with respect to \( u \):**

   \[
   f(u) = \int (u^2 - 2u + 2y) \, du
   \]

   \[
   f(u) = \frac{u^3}{3} - u^2 + 2yu + C(y)
   \]

   where \( C(y) \) is an arbitrary function of \( y \).

6. **Substitute back \( u = x + y \):**

   \[
   f(x, y) = \frac{(x+y)^3}{3} - (x+y)^2 + 2y(x+y) + C(y)
   \]

7. **Use the initial condition \( f(y, y) = -y^3 \) to find \( C(y) \):**

   \[
   f(y, y) = \frac{(y+y)^3}{3} - (y+y)^2 + 2y(y+y) + C(y)
   \]

   \[
   -y^3 = \frac{(2y)^3}{3} - (2y)^2 + 2y(2y) + C(y)
   \]

   \[
   -y^3 = \frac{8y^3}{3} - 4y^2 + 4y^2 + C(y)
   \]

   \[
   -y^3 = \frac{8y^3}{3} + C(y)
   \]

   \[
   C(y) = -y^3 - \frac{8y^3}{3}
   \]

   \[
   C(y) = -\frac{11y^3}{3}
   \]

8. **Substitute \( C(y) \) back into the expression for \( f(x, y) \):**

   \[
   f(x, y) = \frac{(x+y)^3}{3} - (x+y)^2 + 2y(x+y) - \frac{11y^3}{3}
   \]

Therefore, the expression for \( f(x, y) \) is:

\[
f(x, y) = \frac{(x+y)^3}{3} - (x+y)^2 + 2y(x+y) - \frac{11y^3}{3}
\]

Would you like any further details or have any questions?

Here are some related questions to expand on this topic:

1. How do you verify the solution \( f(x, y) \) by substituting it back into the original PDE?
2. What is the role of the arbitrary function \( C(y) \) in finding the particular solution?
3. How does the method of characteristics apply to solving this type of PDE?
4. Can this problem be solved using another method, such as separation of variables?
5. What are the implications of having an initial condition like \( f(y, y) = -y^3 \)?
6. How can we generalize this approach to other types of partial differential equations?
7. What would change in the solution if the initial condition were different, for example, \( f(y, y) = y^3 \)?
8. How do boundary conditions affect the solution of PDEs in general?

**Tip:** Always verify the solution by substituting it back into the original equation to ensure all conditions are satisfied.

Learn how to solve a partial differential equation using the method of characteristics. This detailed step-by-step solution involves transforming the PDE, integrating with respect to the characteristic variable, and applying initial conditions. Suitable for advanced undergraduate students studying partial differential equations.

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