Hamiltonian system; breakdown for different level set values I have a system of differential equations defined by the hamiltonian of the scalar function $H=y^2+e^{-xy}-c$, for some $c>0$. I am asked to describe what happens for $c=1$. I can tell there is a problem using mathematica (the level sets for $c=1$ seem to cross, violating uniqueness of solutions to the ode defined by these partials which travel around level sets).
Since finding a function $H(x(t),y(t))$ in the first place for a system like this would seem to rely on the implicit function theorem, I am looking for a vanishing partial at the value c=1 to show that for this point the ift fails. 
I also tried thinking about the gradient of this function being not surjective for this value of c, since this would imply a singularity (or not regular point), I think. However, I think this would involve plugging the expression for $y(x)$ into my grad function, which seems a bit laborious and i want to make sure the thinking is correct first. 
 A: This ends up being just an application of the implicit function theorem as drawing a level set is equivalent to somehow locally graphing the implicit function of the arguments $x,y$ of $H$ in one dimension lower, as $(x,g(x))$ or $(f(y),y))$ for some suitable functions defined on some open interval. To see this, note $H(x,y)=c$ implies we have graphed $(x,y)=H^{-1}(c)$ in the plane, which only makes sense if we can graph $x$ and $y$ as functions of each other at all points for this particular value of $c$.  
Plugging in $c=1$, we see that $H(x,y)=y^2 +e^{-xy}-1=0$ when $(x,y)=(0,0)$. Part 1 of the implicit function theorem is satisfied. Now I check partials:
$\frac{\partial H}{\partial x}\vert_{(0,0)}=0$ leading to a failure of the implicit function theorem for $c=1$ and the variable $x$. 
I think there is a way to see this by trudging forward naively as well. By trying to solve for $x$ as a function of $y$ near the origin, we get:
$x=\frac{\ln(c-y^2)}{-y}$, which for $c=1$ is $x=\frac{\ln(1-y^2)}{-y}=\frac{\ln((1-y)(1+y))}{-y}=\frac{\ln(1+y)}{-y}+\frac{\ln(1-y)}{-y}$, which around the origin is the same as gives us an indeterminate form which we solve using l'hospital's rule as, taking the limit as $y$ approaches 0: $x=0=1$, a contradiction. 
