Given $f(x,y) = c$, and $C$ is the curve created, we must integrate
$$\int_C ds$$
The parametric way of doing this would be to create functions $x=g(t)$ and $y=h(t)$ to describe the curve. Then we must integrate
$$\int_C \sqrt{\big(g'(t)\big)^2 + \big(h'(t)\big)^2}dt$$
Another way, which may be easier in certain cases, would be to use the Divergence Theorem, as Raffaele points to. Define $\mathcal R = \{(x,y):f(x,y)\le 0\}$ such that $C = \partial R$. Now note that $[f_x,f_y]$ is in the direction of $ds$, so we have
$$\int_C ds = \int_C \frac{[f_x,f_y]}{||[f_x,f_y]||} \cdot ds=\iint_R \left[\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right] \cdot \frac{[f_x,f_y]}{||[f_x,f_y]||} dS $$
where $dS$ is the Area differential. In either case, you must find some expression for the bounds, but one of these integrals may be easier than the other.