Question

How do you implicitly differentiation `1-xy = x-y`?

Answer 1

You have to use the implicit function theorem (aka the Dini's theorem)

The implicit equation is `F(x,y)=1-x-xy+y=0`, F is a smooth function (it's a polynomial), so we can use Dini's theorem

Notation: `F_x={dF}/dx`

Dini's theorem says that in every point `(x,y)` such that `F_y!=0`, we have a neighbourhood where `y=f(x)` with `f` smooth and `f'=-F_x/F_y`

So, `F_y=-x+1 => \forall (x,y) != (1,y)`
`f'(x)=-(-y-1)/(-x+1)=(1+f(x))/(1-x)`

Now we invert, `F_x= -y-1 => forall (x,y)!=(x,-1)`
`x=g(y) and g'(y)=-(1-x)/(-y-1)=(1-g(y))/(1+y)`

So in `(1,-1)` the function is not differentiable, elsewhere is differentiable (it's `C^1`).

Notice that we don't have an "explicit" formula for the derivatives

Answer 2

`1-xy=x-y`

When we differentiate, we'll need the product rule for the second term on the left.
I use the order: `(fg)' = f'g+fg'` for this term we'll have `f=-x` and `g=y`

`d/(dx)(1-xy) = d/(dx)(x-y)`

`0 +d/dx(-xy) = 1 - dy/dx`

`(-1)(y) + (-x)(dy/dx) = 1 - dy/dx`

`-y -x dy/dx = 1 - dy/dx`

So

`dy/dx - x dy/dx = 1+y`

And

`dy/dx = (1+y)/(1-x)`