I'm assuming you want to find #(dy)/(dx)#. For this we first need an expression for #y# in terms of #x#. We note that this problem has various solutions, since #tan(x)# is a periodic functions, #tan(x-y)=x# will have multiple solutions. However, since we know the period of the tangent function (#pi#), we can do the following: #x-y=tan^(-1)x+npi#, where #tan^(-1)# is the inverse function of the tangent giving values between #-pi/2# and #pi/2# and the factor #npi# has been added to account for the periodicity of the tangent.

This gives us #y=x-tan^(-1)x-npi#, therefore #(dy)/(dx)=1-d/(dx)tan^(-1)x#, note that the factor #npi# has disappeared. Now we need to find #d/(dx)tan^(-1)x#. This is quite tricky, but doable using the reverse function theorem.

Setting #u=tan^(-1)x#, we have #x=tanu=sinu/cosu#, so #(dx)/(du)=(cos^2u+sin^2u)/cos^2u=1/cos^2u#, using the quotient rule and some trigonometric identities. Using the inverse function theorem (which states that if #(dx)/(du)# is continuous and non-zero, we have #(du)/(dx)=1/((dx)/(du))#), we have #(du)/(dx)=cos^2u#. Now we need to express #cos^2u# in terms of x.

To do this, we use some trigonometry. Given a right triangle with sides #a,b,c# where #c# is the hypotenuse and #a,b# connected to the right angle. If #u# is the angle where side #c# intersects side #a#, we have #x=tanu=b/a#. With the symbols #a,b,c# in the equations we denote de length of these edges. #cosu=a/c# and using Pythagoras theorem, we find #c=sqrt(a^2+b^2)=asqrt(1+(b/a)^2)=asqrt(1+x^2)#. This gives #cosu=1/sqrt(1+x^2)#, so #(du)/(dx)=1/(1+x^2)#.

Since #u=tan^(-1)x#, we can substitute this into our equation for #(dy)/(dx)# and find #(dy)/(dx)=1-1/(1+x^2)=x^2/(1+x^2)#.