I got $z = x+iy$ and $w=u+iv=\frac1 z = \frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$
$u(x,y) = \frac{x}{x^2+y^2}$ and $v(x,y) = -\frac{y}{x^2+y^2}$
also $x(u,v) = \frac{u}{u^2+v^2}$ and $\;y(u,v) = -\frac{v}{u^2+v^2}$
$x-y<2 \implies \dfrac{u+v}{u^2+v^2} <2$
$x+y<2 \implies\dfrac{u-v}{u^2+v^2} > 2$
How to proceed from here? can't really figure out the region in $w$-plane.
Answering my own question. I would like input/correction form any complex-analysis expert out there.
$\frac{u+v}{u^2+v^2} <2$ when compared to the standard equation of circle: $x^2+y^2+\frac{B}{A}x+\frac{C}{A}y+D = 0$, gives a region in uv- plane which is outside the circle centered at (0.25,0.25) and radius=$\sqrt{2}/4$
Similarly $\frac{u-v}{u^2+v^2} >2$ is the region inside the circle centered at (0.25,-0.25) and radius=$\sqrt{2}/4$
The required region is the intersection of these two regions, which would lie inside the second circle.