Let $x$ be a positive real number such that $x - \frac{1}{x} = 3.$  Find $x + \frac{1}{x}.$
Squaring the equation $x - \frac{1}{x} = 3,$ we get
\[x^2 - 2 + \frac{1}{x^2} = 9.\]Adding 4, we get $x^2 + 2 + \frac{1}{x}^2 = 13,$ so
\[\left( x + \frac{1}{x} \right)^2 = 13.\]Since $x$ is positive,
\[x + \frac{1}{x} = \boxed{\sqrt{13}}.\]