$y'= t$
$y''=t'$
$xt'+2t=0$
$x \large\frac{dt}{dx}$$=-2t$
$\large\frac{dt}{t}=\frac{-2 dx}{x}$
$\log t = -2 \log x +\log c$
$t= \large\frac{c}{x^2}$
$\large\frac{dy}{dx}=\frac{c}{x^2}$
$dy =\large\frac{c dx}{x^2}$
$y= \large\frac{-c}{x}$$+c_1$
It satisfies $(1,1)$
$1=-c+c_1$
Curve intersects the line $y=x$ at right angle means it is normal to curve.
So, $\large\frac{dy}{dx}=\frac{c}{x^2}$$=-1$
$c=-1$
So, $c_1=0$
So, $y= \large\frac{1}{x}$
Hence b is the correct answer.