Differentiate $y^2-2xy+3y=7x$ w.r.t. $x$. Hence show that $\frac{d^2y}{dx^2}(2y-2x+3)=\frac{dy}{dx}(4-2\frac{dy}{dx}).$ 
Differentiate $y^2-2xy+3y=7x$ w.r.t. $x$. Hence show that $\frac{d^2y}{dx^2}(2y-2x+3)=\frac{dy}{dx}(4-2\frac{dy}{dx}).$

I differentiated $y^2-2xy+3y=7x$ w.r.t. $x$ and got:
$$\frac{dy}{dx}=\frac{7+2y}{2y-2x+3}$$
Differentiating one more time, I get:
$$\frac{d^2y}{dx^2}=\frac{-4(x+2)\frac{dy}{dx}+4y+14}{(2y-2x+3)^2}$$
Multiplying by $(2y-2x+3)$ I get:
$$\frac{d^2y}{dx^2}(2y-2x+3)=\frac{-4(x+2)\frac{dy}{dx}+4y+14}{(2y-2x+3)}$$
Now if I understand correctly the question wants me to show that $\frac{d^2y}{dx^2}(2y-2x+3)=\frac{dy}{dx}(4-2\frac{dy}{dx})$
So from what I already got, I guess I need to show that
$$\frac{dy}{dx}(4-2\frac{dy}{dx})=\frac{-4(x+2)\frac{dy}{dx}+4y+14}{(2y-2x+3)}$$
But I don't know how to do this and I'm not sure if I've been following the question correctly. I reckon I might be over-complicating things too. Grateful for any hints/guidance.
 A: Rule of thumb: Try to never differentiate quotients if you can avoid it; they're gross.
You correctly found $$\frac{dy}{dx}=\frac{7+2y}{2y-2x+3}$$
which can equally be written as $$\frac{dy}{dx}(2y-2x+3)=7+2y$$
Differentiating both sides with respect to $x$, the product rule gives $$\frac{d^{2}y}{dx^{2}}(2y-2x+3)+\frac{dy}{dx}\left(2\frac{dy}{dx}-2\right)=2\frac{dy}
{dx}$$
which rearranges to what you want.
A: By starting with $y^{2} - 2 x y + 3 y = 7 x$ then differentiation leads to:
\begin{align}
2 \, y \, y' - 2 \, x \, y' - 2 y + 3 y' = 7.
\end{align}
Differentiating a second time leads to
\begin{align}
2 \, y \, y'' + 2 (y')^{2} - 2 \, x \, y'' - 4 y' + 3 y'' = 0 
\end{align}
The second equation can be reformed into
\begin{align}
(2 y - 2 x + 3) \, y'' = (4 - 2 y' ) \, y'
\end{align}
which is the desired result. 
A: $y^2−2xy+3y=7x$
differentiate both sides wrt $x$,we get
$2y\frac{dy}{dx}-2x\frac{dy}{dx}-2y\frac{dx}{dx}+3\frac{dy}{dx}=7$
$(2y-2x+3)\frac{dy}{dx}=7+2y$
Now without taking $(2y-2x+3)$ into denominator of $7+2y$,differentiate both sides wrt $x$,using product rule
$(2y-2x+3)\frac{d^2y}{dx^2}+\frac{dy}{dx}(2\frac{dy}{dx}-2)=2\frac{dy}{dx}$
$(2y-2x+3)\frac{d^2y}{dx^2}=2\frac{dy}{dx}-\frac{dy}{dx}(2\frac{dy}{dx}-2)$
$(2y-2x+3)\frac{d^2y}{dx^2}=(2-2\frac{dy}{dx}+2)\frac{dy}{dx}$
$(2y-2x+3)\frac{d^2y}{dx^2}=(4-2\frac{dy}{dx})\frac{dy}{dx}$
