Area between the level sets of a curve I have the following curve on the usual $x$-$y$ Euclidean plane:
\begin{align}
(x^2-y^2+2x)^2 = 4\alpha({x^2-y^2})
\end{align}
for some $\alpha > 1$. You may plot the curve here: 
http://www.mathsisfun.com/data/grapher-equation.html
by entering for example, ((x^2)-y^2+2*x)^2=4*12*(x^2-y^2), to check out the case for $\alpha = 12$.
For $\alpha = 6$, say I get one level set, and for $\alpha = 6+\epsilon$ for some small $\epsilon$, I get another level set. I want to show that in the right half plane, the area between the two level sets will be bounded by $\epsilon$ times a constant. Of course, I would like to show that this property holds for any $\alpha$ and sufficiently small $\epsilon$ (In this case the constant multiplier may depend on $\alpha$).
This curve with the equation above is just one example. The main question is how can I do this for any curve (assuming that curve has some certain properties)? (Although it would also be nice to see an answer for the particular case above). Answers in the form "Read this noob" would also be greatly appreciated.
$\textbf{Edit:}$ To clarify, formally, we consider the solutions of the equation $f(x_1,\ldots,x_n) - \alpha g(x_1,\ldots,x_n) = 0$, where $x_1,\ldots,x_n$ are real coordinates, $f$ and $g$ are (say) polynomials. Let $\mathcal{S}_{\alpha} = \{(x_1,\ldots,x_n)\in\mathbb{R}^n:f(x_1,\ldots,x_n) - \alpha g(x_1,\ldots,x_n) = 0\}$. Roughly, I would like to show something like ($\mu$ is the Lebesgue measure)
\begin{align}
\forall\alpha>0,\,\forall\epsilon>0,\,\mu\left(\bigcup_{\beta\in[\alpha,\alpha+\epsilon]}\mathcal{S}_{\beta}\right)\leq h(\alpha)\epsilon,
\end{align}
for some $0<h(\alpha)<\infty$ whenever it is possible to do so. In the definition of $\mathcal{S}_{\alpha}$, we may restrict the solutions to e.g. only positive coordinates as in the example above.
$\textbf{Postmortem:}$ I have got two great answers, none of which however were able to completely resolve the problem. Robert Israel's argument is quite general, but to my reading, does not simply the problem as much as I was expecting. Anıl Başeski's argument is very specific, but it at least solves the particular case as posed in the question. The particular case was related to my own research, and the existence of a parametrization and a closed form solution for the area is quite helpful. I have therefore decided to award the bounty to Anıl. I would also like to thank all who have participated in the dicussion. The general question however is still open, and hence there are no "accepted answers."
 A: If $S_\alpha$ is a compact $n-1$-manifold you should get $h(\alpha) = \int_{S_\alpha} \dfrac{1}{\|\nabla (f/g)\|}$, when that is finite.  A counterexample would be the curve $(x^2 + y^2 - 1)^3 = \alpha$, which is a circle of radius $1 + \alpha^{1/3}$.  The area between $S(0)$ and $S(\epsilon)$ is approximately $2 \pi \epsilon^{1/3}$ and thus is not $O(\epsilon)$ as $\epsilon \to 0$.
A: For your specific equation the curve in the right half plane can be parameterized by $y=t\ x(t)$ to give
$$x(t)=\frac{2\times \sqrt{\alpha (1-t^2)}-2}{1-t^2}$$
$$y(t)=t\ x(t)=t\frac{2\times\sqrt{\alpha (1-t^2)}-2}{1-t^2}$$
where $-\beta\lt t\lt\beta$ and $\beta=\frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}$
The area can be formulated as
$$A=\frac{1}{2}\int_{-\beta}^{\beta}\bigg(x(t)\frac{dy(t)}{dt}-y(t)\frac{dx(t)}{dt}\bigg)dt $$ 
Due to fact that $y=t\ x(t)$
$$A=\frac{1}{2}\int_{-\beta}^{\beta}\bigg(x(t)^2\bigg)dt =F(\beta)-F(-\beta)$$
and
$$F=\frac{t \bigg( 8 \sqrt{\alpha (1-t^2)}-2\bigg ) + 
 2 (1 + 2 \alpha) (-1 + t^2) \text{arctanh}(t)}{t^2-1}$$
which gives following formulation for $A$
$$A=2 (1 + 2 \alpha) \text{arctanh}\bigg(\frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}\bigg)-6\sqrt{\alpha^2-\alpha} $$
By Taylor expansion
$$dA=A(\alpha+\epsilon)-A(\alpha)=\frac{dA}{d\alpha}\epsilon+O(\epsilon^2)$$
and by neglecting higher order terms
$$dA= 4\bigg( \text{arctanh}\bigg(\frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}\bigg)- \frac{\sqrt{\alpha-1}}{\sqrt{\alpha}}\bigg)\epsilon $$
PS: Since equations are too long and time is too short I recommend you to check for typos.
