I'll include more steps than you might want to write when you get more experience, but here we go:
#x^2+2xy-y^2+x=2# Differentiate both sides with respect to #x#.
#d/(dx)(x^2+2xy-y^2+x) = d/(dx)(2)# (Write this every time)
#d/(dx)(x^2)+d/(dx)(+2xy)-d/(dx)(y^2)+d/(dx)(x) = d/(dx)(2)#
Remember that y is the name of some function(s) of #x# that I haven't found expressions for.
So #2xy# is really #2x*("some f of x")# We'll need the product and chain rules.(I use the product rule written as #(FS)' = F'S + FS'#.)
Similarly #y^2# is really #("some f of x")^2# so we'll need the power and chain rules. (Implicit differentiation is using the chain rule.)
#2x (dx)/(dx) + [d/(dx)(+2x)*y+ (+2x) d/(dx)(y)] -2ydy/dx+ 1 = 0#
#2x + 2y+ 2x (dy)/(dx)-2ydy/dx + 1 = 0#
We could solve for #(dy)/(dx) = (-(2x+2y+1))/(2x-2y)#, but if we don't have to, it's usually easier to substitute numbers now:
At #(1,2)#, we get:
#2(1) +2(2) +2(1)(dy)/(dx) -2(2)dy/dx+1 =0# so
#2+4+2(dy)/(dx)-4dy/dx+1=0#
#7-2(dy)/(dx)=0# and finally:
#(dy)/(dx) = 7/2#
The tangent line contains the point #(1, 2)# and has slope #m=7/2# so its equation is:
#y=7/2x-3/2#
With experience, your solution will look more like:
#x^2+2xy-y^2+x=2#
#d/(dx)(x^2+2xy-y^2+x) = d/(dx)(2)#
#2x + 2y+ 2x (dy)/(dx) -2ydy/dx+ 1 = 0#
At #(1,2)#:
#2+4+2(dy)/(dx)-4dy/dx+1=0#
#(dy)/(dx) = 7/2#
