Find all polynomials $P(x)$ with real coefficients such that \[xP\bigg(\frac{y}{x}\bigg)+yP\bigg(\frac{x}{y}\bigg)=x+y\] for all nonzero real numbers $x$ and $y$.
1. **Substitute \( x = 1 \) and \( y = 1 \) into the given equation:**
   \[
   1 \cdot P\left(\frac{1}{1}\right) + 1 \cdot P\left(\frac{1}{1}\right) = 1 + 1
   \]
   Simplifying, we get:
   \[
   2P(1) = 2 \implies P(1) = 1
   \]

2. **Define a new polynomial \( Q(t) \) to simplify the problem:**
   Let \( c = P(0) \). Define \( Q(t) = P(t) - (1-c)t - c \). Then:
   \[
   Q(0) = P(0) - c = 0 \quad \text{and} \quad Q(1) = P(1) - (1-c) - c = 1 - 1 = 0
   \]
   Thus, \( Q(0) = 0 \) and \( Q(1) = 0 \).

3. **Substitute \( P(t) = Q(t) + (1-c)t + c \) into the original equation:**
   \[
   xP\left(\frac{y}{x}\right) + yP\left(\frac{x}{y}\right) = x \left[ Q\left(\frac{y}{x}\right) + (1-c)\frac{y}{x} + c \right] + y \left[ Q\left(\frac{x}{y}\right) + (1-c)\frac{x}{y} + c \right]
   \]
   Simplifying, we get:
   \[
   xQ\left(\frac{y}{x}\right) + yQ\left(\frac{x}{y}\right) + (1-c)y + cx + (1-c)x + cy = x + y
   \]
   \[
   xQ\left(\frac{y}{x}\right) + yQ\left(\frac{x}{y}\right) + x + y = x + y
   \]
   Therefore:
   \[
   xQ\left(\frac{y}{x}\right) + yQ\left(\frac{x}{y}\right) = 0
   \]

4. **Analyze the equation \( xQ\left(\frac{y}{x}\right) + yQ\left(\frac{x}{y}\right) = 0 \):**
   Substitute \( y = 1 \):
   \[
   xQ\left(\frac{1}{x}\right) + Q(x) = 0 \implies Q(x) = -xQ\left(\frac{1}{x}\right)
   \]
   This implies that \( Q(t) \) must be a polynomial such that \( Q(t) = -tQ\left(\frac{1}{t}\right) \).

5. **Determine the degree of \( Q(t) \):**
   Since \( Q(0) = 0 \) and \( Q(1) = 0 \), \( Q(t) \) has roots at \( t = 0 \) and \( t = 1 \). The polynomial \( Q(t) \) must be of the form \( Q(t) = k t(t-1) \) for some constant \( k \).

6. **Verify the form of \( Q(t) \):**
   Substitute \( Q(t) = k t(t-1) \) into \( Q(x) = -xQ\left(\frac{1}{x}\right) \):
   \[
   k x(x-1) = -x k \left(\frac{1}{x}\right)\left(\frac{1}{x} - 1\right)
   \]
   Simplifying the right-hand side:
   \[
   k x(x-1) = -x k \left(\frac{1}{x}\right)\left(\frac{1-x}{x}\right) = -k (1-x)
   \]
   \[
   k x(x-1) = -k (1-x)
   \]
   This equation holds for all \( x \) if and only if \( k = 0 \). Therefore, \( Q(t) = 0 \).

7. **Conclude the form of \( P(t) \):**
   Since \( Q(t) = 0 \), we have:
   \[
   P(t) = (1-c)t + c
   \]
   Using \( P(1) = 1 \):
   \[
   1 = (1-c) \cdot 1 + c \implies 1 = 1
   \]
   This is always true, so the form of \( P(t) \) is:
   \[
   P(t) = t
   \]

The final answer is \( \boxed{P(t) = t} \).