Derivative equals 1 Consider the following curve:
$$
x^{2}+y^{2}=2 x y+16
$$
By differentiating both sides with respect to $x$
$$
\frac{d y}{d x}(2 y-2 x)=2 y-2 x
$$
$$
\frac{d y}{d x}=\frac{2 y-2 x}{2 y-2 x}=1
$$

Are there any reasons that explain why the derivative at each point is equal to 1?

Update:  I think the reason is related to the fact that the curve resembles parallel lines? But how does this explain it?

$$
x^2+y^2=2xy+16\implies\\
x^2-2xy+y^2=16\implies\\
(x-y)^2=16\implies\\
|x-y|=4\implies\\
x>y,\ y=x-4\\
x<y,\ y=x+4.
$$
So, the original implicit function $x^2+y^2=2xy+16$ is equivalent to the two functions $y=x-4$ and $y=x+4$ for different values of $x$ and $y$. Graphically, those are nothing more than two lines of slope $1$. That's why it's only natural to expect the derivative to be $1$.
