Answer
$$f(x)=e^{x+c}~~\text{or}~~f(x)=e^{-(x+c)}$$
where $c \in \mathbb R$.
Work Step by Step
Differentiate both sides:
$$f\int \frac{1}{f}dx+\frac{1}{f}\int fdx=0$$
$$-f\frac{1}{\int fdx}+\frac{1}{f}\int fdx=0$$
$$\frac{1}{f}\int fdx=f\frac{1}{\int fdx}$$
$$\left(\int fdx\right )^{2}=f^{2}$$
$$\int fdx=f~~\text{or}~~\int fdx=-f$$
$$f=f'~~\text{or}~~f=-f'$$
$$\int1 dx=\int\frac{f'}{f}dx~~\text{or}~~\int1 dx=-\int\frac{f'}{f}dx$$
$$x+c=\ln(f)~~\text{or}~~x+c=-\ln(f)$$
$$f(x)=e^{x+c}~~\text{or}~~f(x)=e^{-(x+c)}$$
where $c \in \mathbb R$.
