For $k \geq 0$ we get
$$f(k+2) - f(k+1) = \frac{1}{f(k+1) + f(k+2)}$$
hence $f^{2}(k+2) = f^{2}(k+1) + 1$. It follows
$$4 = f^{2}(4) = f^{2}(3) + 1$$
hence $f(3) = \sqrt{3}$. Also, $3 = f^{2}(3) = f^{2}(2) + 1$ implies
$$f(2) = \sqrt{2} \quad \text{and} \quad 2 = f^{2}(2) = f^{2}(1) + 1$$
gives $f(1) = 1$. Finally, $f(1) = \frac{1}{f(0) + f(1)}$ implies
$$1 = \frac{1}{f(0) + 1}$$
hence $f(0) = 0$.

Now we prove by induction that for any $n \geq 0$, we have $f(n) = \sqrt{n}$. Assume that $f(k+1) = \sqrt{k+1}$, and get
$$f^{2}(k+2) = f^{2}(k+1) + 1 = k+1+1 = k+2$$
hence $f(k+2) = \sqrt{k+2}$ and we are done.