When we find $dy \over dx$ of the equation ${1 \over x} + {1\over y} = x - y$, we can differentiate both sides to obtain: ${dy \over dx} = {y^2(x^2 + 1)\over x^2(y^2-1)}$ ...(1)
On the other hand, we can first transform the equation into $y + x = x^2y - xy^2$, and then differentiate both sides to obtain: ${dy \over dx} = {y^2 -2xy +1 \over x^2 -2xy - 1}$ ...(2)
I have difficulty to derive (1) to (2) or vice versa. I would like to ask for help. Thanks a lot!
We can try to solve it backwards.
Compare $(1)$ and $(2)$:
$$ \frac{y^2(x^2 + 1)}{x^2(y^2 - 1)} = \frac{y^2 - 2xy + 1}{x^2 - 2xy - 1} $$
$$ y^2(x^2 + 1) \cdot (x^2 - 2xy - 1) = x^2(y^2 - 1) \cdot (y^2 - 2xy + 1) \quad \quad (3) $$
If you now calculate each side and cancell common terms you will obtain: $$ y^2x^4 - 2xy^3 - y^2 = x^2y^4 + 2x^3y - x^2 $$ $$ y^2x^4 - 2x^3y + x^2 = x^2y^4 + 2xy^3 + y^2 $$ $$ (yx^2 - x)^2 = (xy^2 + y)^2 $$
Now consider given equation: $$ \frac{1}{x} + \frac{1}{y} = x - y \iff y + x = x^2y - y^2x \iff yx^2 - x = xy^2 + y $$ Thus: $$ \frac{1}{x} + \frac{1}{y} = x - y \iff yx^2 - x = xy^2 + y \implies (yx^2 - x)^2 = (xy^2 + y)^2 $$ Therefore indeed $(1) = (2)$.
If you tried to achieve that equality by forward computation it could prove difficult due to fact of cancellation of a couple of terms somewhere in between which you would have to add in that kind of solution (but now you know exactly what kind of terms are they by considering $(3)$).
Let's look an the second formula for the derivative:
$\frac{dy}{dx}$=$\frac{y^2-2xy+1}{x^2-2xy-1}$.
Transform the numerator first: $y^2-2xy+1$=$y(y-x)-xy+1$.
Replace (y-x) with $(-\frac{1}{x}-\frac{1}{y})$: $-y(\frac{1}{x}+\frac{1}{y})-xy+1$ = $-\frac{y}{x}-1-xy+1$=$-\frac{y}{x}-xy$.
Now let's look and the denominator (do a similar substitution): $x^2-2xy-1$=$x(x-y)-xy-1$=$x(\frac{1}{x}+\frac{1}{y})-xy-1$=$1+\frac{x}{y}-xy-1$=$\frac{x}{y}-xy$.
Let's rewrite the derivative:
$\frac{dy}{dx}$=$\frac{-\frac{y}{x}-xy}{\frac{x}{y}-xy}$.
Multiply both the numerator and denominator by $(-xy)$:
$\frac{dy}{dx}$=$\frac{y^2+x^2y^2}{-x^2+x^2y^2}$=$\frac{y^2(x^2+1)}{x^2(y^2-1)}$, which is the same as your first formula.
