Too implicit

By the Taylor Series of $y = f(x)$ about the point $x = 0$ , we are interested in:

$f(x) = \Sigma_{k=0}^{\infty} \frac{f^{k}(0)}{k!} \cdot (x-0)^k$

For the implicit curve above, we are specifically interested in the point $(x,y) = (0,1)$ . The first derivative of the curve with respect to x gives:

$1 + y' + (2xy^3 + 3y^{2}x^{2}y') + (4x^{3}y^{5} + 5y^{4}x^{4}y') = 0 \Rightarrow 1 + y' = 0 \Rightarrow y' = -1$

Every subsequent derivative of $y'$ is simply zero. Hence the fourth-order Taylor polynomial is expressible as:

$f(x) = 1 + \frac{-1}{1!}x + \frac{0}{2!}x^2 + \frac{0}{3!}x^3$

and the coefficient sum equals: $1 - 1 + 0 + 0 = \boxed{0}.$