Confusion in implicit differentiation steps 
A curve has equation $$x^3-xy+y^2=7$$ 
  a) show that tangent to the curve is horizontal at $(1,3)$
  b) determine $y''$ at $(1,3)$ 

Solution for a)
using product rule, I got this, 

$$3x^2(y+y^2)=0$$ However the teacher got, 
  $$**3x^2-(y+xy')+2yy'=0**$$ then sub $x=1,y=3$ 
  $$3-(3+y')+6y'=0$$
  $$y'(1)=0$$

Solution for b) 

$$(3x^2-xy'-y+2yy')=0$$
  $$**6x-y'-xy''-y'+2(y')^2+2yy''=0**$$
  $$6x-2y'+y''(2y-x)+2(y')^2=0$$
  sub x=1,y=3
  $$6+y''(1)(6-1)=0$$ 
  using y'(1)=0
  $$y''(1)=-\frac{6}{5}$$

I have difficult understanding steps where the stars are marked!
Appreciate your helps! Thanks
 A: Differentiate both sides with respect to $x$. Recall that $y$ is a function of $x$. Maybe it will help if instead of $y$ we write $f(x)$. So your equation becomes
$$x^3-xf(x)+(f(x))^2=7.$$
Now let's differentiate. The derivative of $x^3$ is $3x^2$. 
Now differentiate $xf(x)$. This is a product. So we use the Product Rule, and get $xf'(x)+f(x)$.
Now differentiate the function $(f(x))^2$. You can view this as a product. Then the derivative is $f(x)f'(x)+f(x)f'(x)=2f(x)f'(x)$. But for more complicated expressions, it is better to use the Chain Rule. Our function $(f(x))^2$ is $h(f(x))$ where $h(u)=u^2$. So the derivative is $(2u)f'(x)$, that is, $2f(x)f'(x)$.
On the right-hand side, we have the constant function $7$. Its derivative is $0$.
So we get
$$3x^2 -(xf'(x)+f(x))+2f(x)f'(x)=0.\tag{$1$}$$
Now we go back to the letter $y$. This really should be $y(x)$, but we suppress the $x$. Look back at Equation $(1)$, and write $y$ for $f(x)$ and $y'$ for $f'(x)$. We get
$$3x^2-(xy'+y)+2yy'=0.$$
Now let's do it without the replacement by $f(x)$. The derivative of $x^3$ is $3x^2$. For the derivative of $xy$ with respect to $x$, note that $xy$ is the product of two functions. So we use the Product rule and get $xy'+y$. Finally, to differentiate $y^2$ with respect to $x$, note that by the Chain Rule, $\frac{d}{dx}(y^2)=\frac{d}{dy}(y^2)\frac{dy}{dx}=2y\frac{dy}{dx}=2yy'$. 
A: Hint:
Take the derivative of each term by itself using partial derivatives:
$$x^3-xy+y^2=7$$ 
$x^3 ~\text{partial}~: 3x^2$ (Power Rule)
$-xy ~\text{partial}~: -y -xy'$ (Product Rule)
$y^2 ~\text{partial}~: 2 y y'$ (Power Rule)
Now sum those and you get what the teacher has.
Find partial of: 
$$(3x^2-xy'-y+2yy')=0$$
Again, do the partial of one term at a time:
$3x^2 ~\text{partial}~: 6x$
$-xy' ~\text{partial}~: -y' -xy''$
$-y ~\text{partial}~: -y'$
$2yy' ~\text{partial}~: 2(y')(y') + 2yy''$
Do you see your issues now?
Regards
