Is $e^y+y^2+1=0$ an implicit solution of $(y-1)^2y'=0$? We had this on a test. Apparently I was wrong but I don't understand why. The question was to examine if $y=y(x)$ for which $$e^y+y^2+1=0$$ is a solution of $$(y-1)^2y'=0$$
I argued the relation $e^y+y^2+1=0$ doesn't define a function on any interval and thus it can't be a solution of any differential equation. My line of thinking was that $e^y+y^2+1 \gt 0$.
I'm aware that by differentiating the former we get the latter.
 A: If $y$ is supposed to be a real-valued function, then certainly what you wrote is correct: $$e^y+y^2+1>0,$$ so the equation does not give us a function on any interval. Or rather, the only function $y(x)$ that satisfies the implicit equation is the empty function (that is, its domain its empty, it is not defined anywhere). 
It makes no sense to formally differentiate the empty function. Or rather, the empty function is differentiable at all points of its domain, simply because there are no points in its domain. Thus, it satisfies the given differential equation at all points of its domain. Also, it satisfies $y'\ne y'$ at all points of its domain, or whatever nonsense equation or inequality you pick, for the same reason.
If $y$ is allowed to take complex values, though, then there are functions $y(x)$ that satisfy the given equation, for example the constant functions $$y(x)=-0.2931632\dots\pm i 1.17265\dots$$ A bit trickier is to show that only constant functions satisfy the equation, and therefore $y'=0$, so the given differential equation is satisfied. Or, one can just argue lazily as follows:
It is indeed the case that implicit differentiation of the given equation gives us $(y-1)^2y'=0$. The thing is this: Using the analytic implicit function theorem  (see the statement on the Wikipedia entry, including the remark on regularity), the equation $e^y+y^2+1=0$ (locally) defines $y$ as an analytic function of $x$. The only analytic functions (in fact, the only differentiable functions) that satisfy $(y-1)^2y'=0$ are the constant functions. 
