First, let's think of this as #f(x)=(x-y)^2*y^-1# instead, because it is simpler to use the product rule rather than the quotient rule.
Product Rule: #y'=dy/dx=g'(x)*h(x)+h'(x)*g(x)#
Let #g(x)=(x-y)^2#. Let's compute for #g'(x)#.
#[1]" "g'(x)=2(x-y)# by Power Rule. We need to use Chain Rule after.
#[2]" "g'(x)=2(x-y)*(1-1*dy/dx)#
#[3]" "g'(x)=2(x-y)-[2(x-y)dy/dx]#
Let #h(x)=y^-1#. Let's compute for #h'(x)#.
#[1]" "h'(x)=-1*y^-2# by Power Rule. We need to use Chain Rule.
#[2]" "h'(x)=-y^-2*1*dy/dx#
#[3]" "h'(x)=-y^-2dy/dx#
Now we can solve for #dy/dx#.
#[1]" "dy/dx=g'(x)*h(x)+h'(x)*g(x)#
#[2]" "dy/dx={2(x-y)-[2(x-y)dy/dx]}*y^-1+[-y^-2dy/dx]"⋅"[(x-y)^2]#
#[3]" "dy/dx=2y^-1(x-y)-[2(x-y)dy/dx]y^-1-y^-2dy/dx(x-y)^2#
#[4]" "dy/dx=(2(x-y))/y-[2(x-y)dy/dx]/y-(dy/dx(x-y)^2)/y^2#
Subtract both sides of the equation by the terms containing #dy/dx#.
#[5]" "dy/dx+[2(x-y)dy/dx]/y+(dy/dx(x-y)^2)/y^2=(2(x-y))/y#
Factor out #dy/dx#.
#[6]" "dy/dx(1+[2(x-y)]/y+((x-y)^2)/y^2)=(2(x-y))/y#
#[7]" "dy/dx((y^2+2(x-y)y+(x-y)^2)/y^2)=(2(x-y))/y#
#[8]" "dy/dx=(2(x-y))/y(y^2/[y-(x-y)]^2)#
#[9]" "color(blue)(dy/dx=(2(x-y)y)/x^2)#
Note: Your solution doesn't actually have to be this long. In fact, you can complete this whole problem in maybe two or three lines. I just broke it down into several parts so you know what I'm doing.
