Find the quadratic approximation of the following functions at the given points

In general the quadratic approxmiation of a function $f(x,y)$ near a points $(a,b)$ is
\[P_2(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\]
\[+\frac{1}{2}f_{xx}(a,b)(x-a)^2+\frac{1}{2}f_{yy}(a,b)(y-a)^2+f_{xy}(a,b)(x-a)^2\]

a) We have $(a,b)=(0,0)$ and 
\[f_x=ye^{xy},  f_y=xe^{xy},  f_{xx}=y^2 e^{xy}, f_{yy}=x^2 e^{xy}, f_{xy}=e^{xy}+xye^{xy}.\]
Hence 
\[f(0,0)=1, f_x(0,0)=0, f_y(0,0)=0, f_{xx}(0,0)=0, f_{yy}(0,0)=0, f_{xy}(0,0)=1.\]
Hence the quadratic approximation is
\[P_2(x,y)=1+1(x-0)(y-0)=1+xy.\]

(b) $g(x,y)=\sinh (xy)$, $(a,b)=(0,0)$, and 
\[g_x=y\cosh (xy), g_y=x\cosh (xy), \]
\[  g_{xx}=y^2\sinh (xy),  g_{yy}=x^2\sin (xy),   g_{xy}=\cosh (xy)+xy\sin h (xy).\]
Hence 
\[g(0,0)=0, g_x(0,0)=0, g_y(0,0)=0,  g_{xx}(0,0)=0, g_{yy}(0,0)=0, g_{xy}(0,0)=1.\]
Thus
\[P_2(x,y)=1(x-0)(y-0)=xy.\]