Implicitly differentiating $y = y^2 x$ I am self learning calculus and have a problem that may be simple here but  I cant find an answer on the web so here it is:
If we have the equation:
$$y = y^2x$$
and differentiate it (implicitly and using product rule) we get:
$$\frac{dy}{dx} = \frac{y^2}{1-2xy}.$$
However, $y =y^2x$ can be simplified to $1 = xy$, i.e. $y = \frac{1}{x} $, which when differentiated using the power rule gives $\frac{dy}{dx} = -\frac{1}{x^2}$. The two differentials are different, as
$\frac{dy}{dx} = -\frac{1}{x^2}$ does not equal $\frac{dy}{dx} = \frac{y^2}{1-2xy}$.
Why is this when we are essentially differentiating the same equation?
(I plotted the graph on wolfram alpha and they are indeed not the same.)
 A: Remember that you found that $y=\frac{1}{x}$.  So, $$\frac{y^2}{1-2xy}=\frac{1/x^2}{1-2}=-\frac{1}{x^2}$$ and the two answers you have found actually are the same.
However, there are some issues with division by $0$ in your second approach.  It is not correct to simplify $y=y^2x$ to $1=xy$; that is only valid if $y\neq 0$.  So in fact, the graph of the equation $y=y^2x$ consists of the parabola $y=\frac{1}{x}$ (along which $\frac{dy}{dx}$ is indeed $-\frac{1}{x^2}$) together with the line $y=0$ (since if $y=0$ the equation $y=y^2x$ is always true).  At points on $y=0$, your first answer $\frac{dy}{dx}=\frac{y^2}{1-2xy}$ is correct (and it simplifies to $0$) but your second answer $\frac{dy}{dx}=-\frac{1}{x^2}$ is wrong.
A: The equation is
$$
y(1-xy)=0
$$
so the curve is the union of the line $y=0$ and of the hyperbola $xy=1$.
The function is locally invertible at every point; the derivative at points with nonzero $y$-coordinate is $y'=-1/x^2$; the derivative at points with zero $y$-coordinate is $y'=0$.
If you do implicit differentiation, you get
$$
y'=2yy'x+y^2
$$
that indeed gives
$$
y'=\frac{y^2}{1-2xy}
$$
which is not contradictory: if $y=0$, then $y'=0$; if $y\ne0$, then $xy=1$ and you get
$$
y'=-y^2=-\frac{1}{x^2}
$$
