# x^2/(e^(x-y))=C #

# :. x^2=Ce^(x-y) #

# :. x^2=Ce^xe^-y #

Differentiating implicitly involves differentiating everything wrt #x# and using the chain rule and product rule:

So, # d/dx(x^2)=d/dx(Ce^xe^-y) #

# :. 2x=Cd/dx(e^xe^-y) #

# :. 2x=C(e^xd/dxe^-y+e^-yd/dxe^x) # using the product rule

# :. 2x=C(e^xd/dye^-ydy/dx+e^-ye^x) # using the chain rule

# :. 2x=C(-e^xe^-ydy/dx+e^-ye^x) #

# :. 2x=x^2/(e^(x-y))(-e^xe^-ydy/dx+e^-ye^x) #

We don't need an explicit expression for #dy/dx# so don't attempt to simplify further. We can just substitute #x=-2# and #y=1#, to find #dy/dx# at that value.

So, at # (-2,1) # we have:

# :. 2x=x^2/(e^(x-y))(-e^xe^-ydy/dx+e^-ye^x) #

# :. 2(-2)=(-2)^2/(e^(-2-1))(-e^-2e^-1 dy/dx+e^-1e^-2)#

# :. -4=4/(e^-3)(-e^-3 dy/dx+e^-3)#

# :. -4=4(- dy/dx+1)#

# :. - dy/dx+1=-1#

# :. dy/dx=2#

So the tangent passes through #(-2,1)# and has slope #m=2# so we use the formula #y-y_1=m(x-x_1)# to get the equation:

# y-1=2(x--2) #

# :. y-1=2(x+2) #

# :. y-1=2x+4 #

# :. y=2x+5 #