We have: $$(x-y)^2 = x+y-1$$
To solve, I added the derivative operator with respect to $x$ on both sides of the equation, which simplified to:
$$(2y-2x)\frac{\mathbb{d}y}{dx} = -2x + 2y + 1 + \frac{\mathbb{d}y}{dx}$$
Here is where I am confused, to solve for $\frac{\mathbb{d}y}{dx}$ we subtract $\frac{\mathbb{d}y}{dx}$ from both sides, I thought the $\frac{\mathbb{d}y}{dx}$ would cancel out on both sides but really what happens is:
$$(2y-2x-1)\frac{\mathbb{d}y}{dx} = -2x + 2y + 1$$
So my question is, why didn't the $\frac{\mathbb{d}y}{dx}$ cancel out and instead became a factored $-1$?
Note that $\frac{dy}{dx} = 1\frac{dy}{dx}$, so when we subtract $\frac{dy}{dx}$ from each side of the equation, we obtain $-\frac{dy}{dx} = -1\frac{dy}{dx}$ on the left-hand side of the equation, which allows us to factor out a $-1$. Details follow.
If we assume that $y$ is a differentiable function of $x$, differentiating both sides of the equation $$(x - y)^2 = x + y - 1$$ implicitly with respect to $x$ yields \begin{align*} 2(x - y) \cdot \frac{d}{dx}(x - y) & = 1 + \frac{dy}{dx} && \text{Chain Rule}\\ 2(x - y)\left(1 - \frac{dy}{dx}\right) & = 1 + \frac{dy}{dx}\\ 2(x - y) - 2(x - y)\frac{dy}{dx} & = 1 + \frac{dy}{dx}\\ 2x - 2y - 2x\frac{dy}{dx} + 2y\frac{dy}{dx} & = 1 + \frac{dy}{dx}\\ -2x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} & = 1 - 2x + 2y\\ (-2x + 2y - 1)\frac{dy}{dx} & = 1 - 2x + 2y\\ \frac{dy}{dx} & = \frac{1 - 2x + 2y}{-2x + 2y - 1}\\ \end{align*}