`y^2-xy=2`

To determine `(dy)/d(x)` , let's apply implicit differentiation. So,

`d/(dx)(y^2-xy = 2)`

`d/(dx) (y^2) - d/(dx) (xy) = d/(dx) (2)`

Note that the derivative of a constant is zero ( (c)' = 0) .

`d/(dx) (y^2) - d/(dx)(xy) = 0`

To take the derivative of `y^2` , apply the power formula which is `(u^n) = n*u^(n-1)u'` .

`2y y' - d/(dx) (xy) = 0`

Also, for `d/(dx) (xy)` , use the product formula of derivatives which is

`(uv)'= vu' + uv'`

`2yy' - (yx' + xy') = 0`

Note that `x' = d/(dx) (x) = 1` . And, `y'= d/(dx)(y)=(dy)/(dx)` .

`2y(dy)/(dx) - (y + x(dy)/(dx))=0`

`2y(dy)/(dx) - y -x(dy)/(dx)=0`

Isolate the terms with `(dy)/(dx)` .

`2y(dy)/(dx)-x(dy)/(dx)=y`

`(dy)/(dx)(2y-x)=y`

`(dy)/(dx)=y/(2y-x)`

**Hence, `(dy)/(dx)=y/(2y-x)` .**