Apply the linear approximation
f(0.01,1.05)≈f(0,1)+fx(0,1)(0.01)+fy(0,1)(0.05),
with
f(x,y)fx(x,y)fy(x,y)=sin(πxy+lny)=πycos(πxy+lny)=(πx+y1)cos(πxy+lny)f(0,1)fx(0,1)fy(0,1)=sin0=0=πcos0=π=cos0=1
This gives
f(0.01,1.05)≈f(0,1)+fx(0,1)(0.01)+fy(0,1)(0.05)=0+π(0.01)+1(0.05)=0.01π+0.05≈0.0814