If the point on the curve $\mathrm{y}-{\mathrm{e}}^{\mathrm{xy}}+\mathrm{x}=0$  at which we have vertical tangent is (a,b) then a+b=________

Curve $\mathrm{y}-{\mathrm{e}}^{\mathrm{xy}}+\mathrm{x}=0$ 
Diff w. r.t.’x’ we get
$\mid \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1+\mathrm{y}\cdot {\mathrm{e}}^{\mathrm{xy}}}{1-{\mathrm{xe}}^{\mathrm{xy}}}$
Let (a,b) is a point on the curve at which we have vertical tangent, so 
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(\mathrm{a},\mathrm{b})}=\mathrm{\infty }\left(\mathrm{or}\right){\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}_{(\mathrm{a},\mathrm{b})}=0$
Which is possible only if a=1 and b=0
$\therefore \text{ Point is }(\mathrm{a},\mathrm{b})=(1,0)\Rightarrow \mathrm{a}+\mathrm{b}=1+0=1$