Where have I gone wrong in trying to solve this ODE? I'm trying to solve: $\frac{dy}{dx}=\frac{x+y-1}{x+4y+2}$.
Attached is a picture of my working.
Could someone please  tell me where I'm going wrong?

I'm tried both Maple and Wolfram and neither of them gives me a 'nice' answer.
I know it's wrong as I've implicitly differentiated my answer and I get the wrong algebraic value of $\frac{dy}{dx}$. Thanks.
 A: One trick to solve such problems somewhat explicitly is to change the parameter in such a way that the denominator is canceled.
That is, set $x=X(t)$ and $y=Y(t)=y(X(t))$ with $\dot X=X+4Y+2$ on maximal intervals where this expression is different from zero. Then
$$\dot Y(t)=\frac{dy}{dx}(X(t))\dot X(t)=\frac{x+y-1}{x+4y+2}(x+4y+2)=X+Y-1$$
and the new system is linear,
$$\begin{bmatrix}\dot X(t)\\\dot Y(t)\end{bmatrix}=
\begin{bmatrix}1&4\\1&1\end{bmatrix}
\begin{bmatrix}X(t)\\Y(t)\end{bmatrix}+
\begin{bmatrix}2\\-1\end{bmatrix}
$$
The characteristic polynomial of the system matrix is $(λ-1)^2-4=(λ+1)(λ−3)$
with eigenvectors $\begin{bmatrix}2\\-1\end{bmatrix}$ for $λ=-1$ and $\begin{bmatrix}2\\1\end{bmatrix}$ for $λ=3$. 
Multiplying from the left with the inverse of the eigenvector matrix (leaving out the determinant) gives
$$
\frac{d}{dt}
\begin{bmatrix}X(t)-2Y(t)\\X(t)+2Y(t)\end{bmatrix}
=
\begin{bmatrix}-X(t)+2Y(t)\\3X(t)+6Y(t)\end{bmatrix}
+
\begin{bmatrix}4\\0\end{bmatrix}
$$
so that $X(t)-2Y(t)=(X(0)-2Y(0)+4)e^{-t}-4$ and $X(t)+2Y(t)=(X(0)+2Y(0))e^{3t}$.
Elimination of $e^t$ from these equations results in the implicit 4th degree equation already derived,
$$
(X(0)+2Y(0))(X(0)-2Y(0)+4)^3=(X(t)+2Y(t))(X(t)-2Y(t)+4)^3
$$ 
and in the original parametrization
$$
(x_0+2y_0)(x_0-2y_0+4)^3=(x+2y(x))(x-2y(x)+4)^3
$$
One can just as well solve for $X(t)$ and $Y(t)$ to obtain formulas for the trajectories,
\begin{align}
X(t)&=\tfrac12(X(0)-2Y(0)+4)e^{-t}-2+\tfrac12(X(0)+2Y(0))e^{3t}\\
Y(t)&=-\tfrac14(X(0)-2Y(0)+4)e^{-t}+1+\tfrac14(X(0)+2Y(0))e^{3t}
\end{align} 
The vertical turning points of this parametrisation, where the solutions for $y(x)$ end, are solutions of
$$\begin{aligned}
0&=2X+8Y+4=3(X+2Y)-(X-2y)+4
\\
&=3(X(0)+2Y(0))e^{3t}-(X(0)-2Y(0)+4)e^{-t}+8
\end{aligned}$$
again a 4th degree polynomial in $e^t$.
A: To continue on from my comment (and losing the absolute value signs for the moment, since we are taking fourth roots - but note the fourth roots vanish in the calculation) your complicated expression can have fractions cleared to give:
$$(x-2y-4)^{\frac 34}(x+2y)^{\frac 14}=\frac 1C$$
Implicit differentiation then gives
$$\frac 34(1-2\frac{dy}{dx})(x-2y-4)^{-\frac 14}(x+2y)^{\frac 14}+\frac 14(1+2\frac{dy}{dx})(x-2y-4)^{\frac 34}(x+2y)^{-\frac 34}=0$$ Which simplifies nicely to $$3(1-2\frac{dy}{dx})(x+2y)=-(1+2\frac{dy}{dx})(x-2y-4)$$ and gathering the terms together then gives:$$3x+6y+x-2y-4=(6x+12y-2x+4y+8)\frac {dy}{dx}$$ whence $$\frac {dy}{dx}=\frac{x+y-1}{x+4y+2}$$
as required (apart from being careless about signs).
