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).
