Substitution(Enjoy!)

Let's write $u = x^2 - y^2 \\ v = \tanh^{-1}\left(-\frac{y}{x}\right)$ then, $\frac{1}{2} du = x\:dx-y\:dy \\ u\: dv = y\: dx-x \:dy$ Hence, we can write the original equation as, $\frac{-du}{2u\:dv} = \sqrt{\frac{1+u}{u}} \:,$ which is now a separable. Now, we have $\frac{du}{2\sqrt{u(1+u)}} = -dv \:.$ Integrating both sides, we get $\ln(\sqrt{u} + \sqrt{1+u} ) = - v + C \:.$ Subtituting back to original $x$ and $y$ , we get (use identity $\tanh^{-1}\left(-\frac{y}{x}\right) = \frac{1}{2} \ln\left(\frac{x-y}{x+y}\right)$ ) $\sqrt{x^2 - y^2} + \sqrt{1+ x^2 - y^2} = c\left(\frac{x+y}{\sqrt{x^2 - y^2}}\right)$ So, we have $f(x,y) = x^2 - y^2$ , and $60 f(3,2) = 300$ .

substitute x=p+q,y=p-q hence x^2-y^2 which implies d(pq)= 2xdx - 2ydy then find d(y/x) using quotient rule. Remember dp/p - dq/q = d(ln(p/q)) . Just substitute in the equation and solve.You will get f(x,y) = x^2-y^2 . Hence 60(3,2)=300