1. **State the problem:** Solve for the real value of $x$ in the equation $$\frac{1}{x+r} = \frac{r}{x^2 - r^2}.$$\n\n2. **Recall the difference of squares formula:** $$x^2 - r^2 = (x-r)(x+r).$$ This will help simplify the denominator on the right side.\n\n3. **Rewrite the equation using the factorization:** $$\frac{1}{x+r} = \frac{r}{(x-r)(x+r)}.$$\n\n4. **Multiply both sides by $(x+r)$ to eliminate the denominator on the left:** $$1 = \frac{r}{x-r}.$$\n\n5. **Multiply both sides by $(x-r)$ to clear the denominator on the right:** $$(x-r) = r.$$\n\n6. **Solve for $x$:** $$x - r = r \implies x = 2r.$$\n\n7. **Check for restrictions:** The original denominators cannot be zero, so $x \neq -r$ and $x^2 \neq r^2$ (i.e., $x \neq \pm r$). Since $x=2r$ does not violate these, it is valid.\n\n**Final answer:** $$\boxed{x = 2r}.$$