How far can I rewrite an equation before the derivative is not equivalent? I've run into a problem that I can't explain to my class. 
We are looking at the derivative for the equation $\frac{x}{y}+\frac{y}{x}=3y$. We calculated it to be $\frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}$ and we also verified it with Wolfram Alpha.
A student thought about rewriting the original equation as $x^2+y^2=3xy^2$ by multiplying everything by $xy$. I realize this adds to the domain and adds the point $(0,0)$ as a solution, but it's not differentiable at that point regardless as it's not continuous at that point. When we took the derivative of the rewritten equation and got $\frac{3y^2-2x}{2y-6xy}$, which is not equivalent to our previous calculation.
I can't figure out why the derivatives are so vastly different if all that was added to the original was a new point that's not differentiable to begin with. 
Any help is much appreciated!
 A: Along a curve $f(x,y) = 0$ the derivative $y'$ is given by
$$
y_f'(x,y) 
= -\frac{\partial_x f(x,y)}{\partial_y f(x,y)}
.
$$
Given some smooth function $\varphi$ which doesn't vanish on the curve, we can create a new function $g(x,y) = \varphi(x,y) \, f(x,y)$ that also vanishes along the curve. For this we get
$$\begin{align}
y_g'(x,y) 
&= -\frac{\partial_x g(x,y)}{\partial_y g(x,y)}
= -\frac{\partial_x \varphi(x,y) \, f(x,y) + \varphi(x,y) \, \partial_x f(x,y)}{\partial_y \varphi(x,y) \, f(x,y) + \varphi(x,y) \, \partial_y f(x,y)} \\
&= -\frac{\partial_x \varphi(x,y) \cdot 0 + \varphi(x,y) \, \partial_x f(x,y)}{\partial_y \varphi(x,y) \cdot 0 + \varphi(x,y) \, \partial_y f(x,y)} 
= -\frac{\partial_x f(x,y)}{\partial_y f(x,y)}
= y_f'(x,y)
\end{align}$$
as expected.
In the given case we have $f(x,y) = \frac{x}{y}+\frac{y}{x} - 3y$, $\varphi(x,y) = xy$ and $g(x,y) = x^2+y^2-3xy^2$. With the help of the above calculations I manage to show that the two expressions are equivalent along the curve:
$$\begin{align}
\frac{3y^2-2x}{2y-6xy}
&= - \frac{y\left(\frac{x}{y}+\frac{y}{x} - 3y\right) + xy\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x\left(\frac{x}{y}+\frac{y}{x} - 3y\right) + xy\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)} \\
&= - \frac{y \cdot 0 + xy\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x \cdot 0 + xy\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)} 
= - \frac{\frac{1}{y}-\frac{y}{x^2}}{-\frac{x}{y^2}+\frac{1}{x}-3} \\ 
&= - \frac{x^2y^2\left(\frac{1}{y}-\frac{y}{x^2}\right)}{x^2y^2\left(-\frac{x}{y^2}+\frac{1}{x}-3\right)} = - \frac{x^2y-y^3}{-x^3+xy^2-3x^2y^2} \\ 
&= \frac{y(x^2-y^2)}{x(3xy^2+x^2-y^2)}
.
\end{align}$$
A: I got the second derivative like:
$$ \frac{x}{y}+\frac{y}{x}= 3y,\,$$
or
$$ \frac{x}{y^2}+\frac{1}{x}= 3$$
Differentiate with quotient rule to obtain
$$ \frac{y^2-2xyy'}{y^4}= \frac{1}{x^2}\,$$
Simplifying
$$ y'=\frac{y(x^2-y^2)}{2x^3}$$
Now comparing only the denominator of your first derivative ( as numerator is same )
$$ 2x^3=x (3x y^2-x^2-y^2)\rightarrow x^2+y^2-3xy^2=0$$ 
which tallies with given equation when multiplying out everything by $xy.$
We can apply L'Hospital's rule thrice if required to evaluate the derivative limit at $(0,0):$ 
$$ {u/v}={u'/v'}={u''/v''}={u'''/v'''}. $$
