Final Answer: #dy/dx={-x^2}/{1+x^2}#
Let #z =(x+y)#
#tan(x+y)=x \quad \rarr \quad tan(z)=x#
Take the derivative of each side with respect to x
#d/dx[tan(z)]=d/dx[x]#
#sec^2(z)dz/dx=1 \quad \larr \quad \text{(Chain Rule)}#
Now find what #dz/dx # is
#dz/dx=d/dx[x+y]=d/dx[x]+d/dx[y]=1+dy/dx#
substitute #z# and #dz/dx# into: #\quad sec^2(z)dz/dx=1# and get
#sec^2(x+y)(1+dy/dx)=1#
Solve for #dy/dx#
#dy/dx=cos^2(x+y)-1=-sin^2(x+y)#
I used the trig identity: #cos^2(z)-1=-sin^2(z)# in the last step.
Substitute #y# to get the answer purely in terms of #x#
#tan(x+y)=x \quad \rarr \quad y=arctan(x)-x#
#dy/dx=-sin^2(arctan(x))#
Answer can be simplified with some work
Let #theta=arctan(x) \implies tan(theta)=x/1=\text{opposite}/\text{adjacent}#
The above expression implies the triangle below
Sub #arctan(x)=theta# into the #dy/dx# expression
#dy/dx=-sin^2(theta)=-(\text{opposite}/\text{hypotenuse})^2#
From the diagram we know #sin(theta)=x/\sqrt{1+x^2}#. So...
#dy/dx={-x^2}/{1+x^2}#
