Need to Find $y'''(1)$ if $x^2 + xy + y^3 = 1$

Question

Yes I'm aware I posted this but I did before adding all these details and I can't wait for it to go off hold... This is the question I cannot seem to figure out how to do. If $x^2 + xy + y^3 = 1$, find the value of $y'''$ at the point where $x = 1$.

I need help with this because my teacher barely speaks English and doesn't answer questions.. Therefore I'm having a lovely time attempting the homework. What I think I've been able to do so far is take the first derivative then setting $x=1$ $$ y' =\frac {-2} {y+3y^2} $$

Then trying to take the second derivative using the quotient rule.. $$ y'' = \frac {(y+3y^2)(0)-(-2)(y+3y^2)'} {(y+3y^2)^2} $$ $$ (y+3y^2)' = (1)\left(\frac{dy}{dx}\right)+6y\left(\frac{dy}{dx}\right) = \frac{-2}{y+3y^2}+\frac{-12y}{y+3y^2} = \frac{-2-12y}{y+3y^2}$$ $$ y''=\frac{0-(-2)\left(\frac{-2-12y}{y+3y^2}\right)}{(y+3y^2)^2} = \frac {-\frac {4+24y} {y+3y^2}}{(y+3y^2)^2}=-\frac {4+24y} {y+3y^2}*\frac{1}{(y+3y^2)^2}$$ $$ y'' = \frac {4+24y} {(y+3y^2)^3} $$ At this point, I'm not sure it this is right.. But I went and tried for the third with more quotient rule! $$y'''= \frac{(y+3y^2)^3(24)\left(\frac{dy}{dx}\right)-(4+24y)\left((y+3y^2)^3\right)'}{(y+3y^2)^6}$$ Before I can finish that problem I need the derivative of $(y+3y^2)^3$... I believe for this I need to use chain Rule.
if $f(x)=x^3$ then $f'(x)=3x^2$ If we then let $g(x) = y+3y^2$ then $g'(x) = 1+6y$ leaving us with... $$3(y+3y^2)^2*(1+6y)$$ Plugging that back in to the rest of the problem we get... $$y'''= \frac{(y+3y^2)^3(24)\left(\frac{dy}{dx}\right)-(4+24y)\left(3(y+3y^2)^2*(1+6y)\right)\left(\frac{dy}{dx}\right)}{(y+3y^2)^6}$$ factoring out a $(y+3y^2)^2$ I get $$y'''=\frac{(y+3y^2)^2\left(y+3y^2)(24)\left(\frac{dy}{dx}\right)-(4+24y)\left(3(1+6y)\right)\left(\frac{dy}{dx}\right)\right)}{(y+3y^2)^6}$$ canceling out for... $$y'''=\frac{\left(y+3y^2)(24)\left(\frac{dy}{dx}\right)-(4+24y)\left(3(1+6y)\right)\left(\frac{dy}{dx}\right)\right)} {(y+3y^2)^4}$$ $$y'''=\frac{24y(1+3y)\left(\frac{dy}{dx}\right)-12(36y^2+12y+1)\left(\frac{dy}{dx}\right)} {(y+3y^2)^4}$$ $$y'''=\frac{\frac{-48y(1+3y)}{y+3y^2}+\frac{24(36y^2+12y+1)}{y+3y^2}} {(y+3y^2)^4}=\frac{\frac{-48y(1+3y)+24(36y^2+12y+1)}{y+3y^2}} {(y+3y^2)^4}$$ Then finally... $$y'''=\frac{-48y(1+3y)+24(36y^2+12y+1)}{y+3y^2}*\frac{1}{(y+3y^2)^4}$$ $$y'''=\frac{-48y(1+3y)+24(36y^2+12y+1)}{(y+3y^2)^5} $$

Answer 1

You have $x^2 + xy + y^3 = 1$ and want $y'''(1)$. Differentiating, we have: $$2x+y+xy'+3y^2y' = 0$$ Again: $$2+y'+y'+xy''+6y(y')^2+3y^2y'' = 0 \implies \\\implies 2 + 2y'+6y(y')^2+(x+3y²)y'' = 0$$ Yet again: $$2y''+6(y')^3+12yy'y''+(1+6yy')y''+(1+3y^2)y''' = 0$$

If $x = 1$, then $x^2+xy+y^3 = 1$ gives $y+y^3 = 0 \implies y(y^2+1) = 0 \implies y = 0$. In the first step, we get $$2+y' = 0 \implies y' = -2$$ Now we sub. $x = 1, y = 0, y' = -2$ in the second step, to get: $$2-4+y'' = 0 \implies y'' = 2$$ Plugging all in the last step: $$4-48+2+y''' = 0 \implies y''' = 42.$$

Answer 2

Differentiating $x^2 + xy + y^3 = 1$ gives $$2x + xy' + y + 3y^2y' = 0$$ Differentiating again gives $$2 + xy'' + 2y' + 6y(y')^2 + 3y^2 y'' = 0$$ A third time gives $$xy''' + 3y'' + 6(y')^3 + 18 y y' y'' + 3y^2y''' = 0$$ In the top equation, settign $x = 1$ and $y = 0$ gives $$2 + y'= 0$$ Which solves to $y' = -{2}$. Inserting into the second equation gives $$2 + y'' - 4 = 0$$ This solves to $y'' = 2$. Putting into the third equation gives $$y''' + 6 - 48 = 0$$ This solves to $y''' = 42$.