Find the 2nd order Taylor Polynomial of y(x) about x = 0, given: I am given the equation:
$$y^3 + y^2 + y - 3 = x$$ 
with $y(0)=1$. I am wondering if what I have done is valid, given that this is a homework question for an Integral Calculus class, but I seem to be able to answer the question without doing any integration.
Taking the derivative of both sides (and using implicit differentiation) with respect to $x$ and then rearranging, I have
$$\frac{d}{dx} \left(y^3 + y^2 + y - 3\right) = \frac{d}{dx}(x)$$
$$\frac{dy}{dx} = \frac{1}{3y^2 + 2y + 1}$$
So, since I am given $y(0) = 1$, I have my first two terms in the Taylor Series: f(0) = 1, and now f'(0) = 1/6, since y = 1 when x = 0.
I then take the 2nd derivative with respect to x (and again using implicit differentiation) to get:
$d^2y/dx^2 = - y'(x)(6y+2)/(3y^2 + 2y + 1)^2$
Plugging in for y and y', when x = 0, and I get y''(0) = -1/27
Therefore my Taylor polynomial of order 2, about x = 0, is:
$1 + x/6 - x^2/27$
Is this correct? 
I appreciate any tips and advice!
 A: Minor error, recall that we need to divide by $2!$ at $x^2$ (and so on). The Taylor expansion about $x=a$ up to the $(x-a)^n$ term goes like this:
$$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+ \cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n.$$
You got after differentiating once,
$$(3y^2+2y+1)\frac{dy}{dx}=1.$$
To find the second derivative, I prefer not to divide, it sometimes makes a mess. The above expression  looks nice, might as well just differentiate as it stands. We get 
$$(3y^2+2y+1)\frac{d^2y}{dx^2}+ (6y+2)\left(\frac{dy}{dx}\right)^2=0,$$
and now we can find the second derivative at $0$. It does indeed simplify to $-1/27$.
A: Alternative:
As a first approximation, let $y=a$ be a constant and match it to the constant term in $x$:
$$a^3+a^2+a-3\to0.$$
We have the solution $a=1$.
Then try a linear approximation and identify the first order terms:
$$(bx+1)^3+(bx+1)^2+(bx+1)-3=\cdots+3bx+2bx+bx\cdots\to x,$$giving $b=\frac16$.
Then add a second degree term, and
$$(cx^2+\frac x6+1)^3+(cx^2+\frac x6+1)^2+(cx^2+\frac x6+1)-3=\\\cdots+3cx^2+\frac{x^2}{12}+2cx^2+\frac{x^2}{18}+cx^2\cdots\to0x^2,$$or $c=-\frac1{54}$.
$$y\approx1+\frac x6-\frac{x^2}{54}\cdots.$$
This coincides with the Taylor development.
