By Taylor method

$$y=y_0+hy'_0+\dfrac {h^2}{2!}y"_0+\dfrac {h^3}{3!}y"'_0 +....$$

$Let y'=\dfrac {dy}{dx}=x^2y-1\\ y"=x^2y'+y2x\\ y'''=x^2y''+4y'x+2y\\ y^{iv}=x^2y'''+y''.2x+y'4+4xy'' +2y'\\ =x^2y''' +6xy''+6y'\\ At \space x_0=0,y_0=1,h=0.1\\ y_0^1=x_0^2y_0-1=0-1=-1\\y''_0=x_0^2y_0-1=0-1=-1\\ y'''_0=x_0^2y''_0+4x_0y'_0+2y_0\\ 0+0+2(1);=2\\ y^{iv}_0=x_0^2+y'''_0+6x_0y''_0+6y'_0\\ =0+0+2(1)\\ =2\\ \text{ Now h=x and using above values in taylor series }\\ y=1+x(-1)+\dfrac {x^2}{2!}(0)+\dfrac {x^3}{3!}(2)+\dfrac {x^4}{4!}(-6) +....\\ y=1-x+ \dfrac {x^3}{3!} -\dfrac {x^4}{4!} + .... (1)\\ put\space x=0.1\\ y=1-0.2+\dfrac{0.2^3}3-\dfrac {(0.2)^4}4+....\\ =0.80227$