Given that #x# and #y# are related by #(x-y)(x+y)-xy+y= C # and that the point #(-1,-1)# lies on the graph of the equation, we could find #C# and the solve for #y# and differentiate.

That looks like a lot of work. It will probably be easier to differentiate implicitly.

Differentiating the expression #(x-y)(x+y)# involves the product rule and that is needlessly complicated. We can expand this product of a sum and difference to get

#x^2-y^2-xy+y=C#.

Differentiating, (using the product rule for #-xy#) we get

#2x-2y dy/dx -y - x dy/dx + dy/dx = 0#

Rather than solve for #dy'dx# before substituting, I find it simpler to substitue the values #x=-1# and #y=-1#, then solve for #dy/dx#.

#2(-1) -2(-1) dy/dx -(-1) - (-1)dy/dx + dy/dx = 0#

#-2+2dy/dx+1+dy/dx+dy/dx = 0#

#-1+4dy/dx=0#

#dy/dx = 1/4#