I need to write Taylor expansion term, for $n=4, k=3$. At a point $M=(1;\pi)$ for the function: $f(x,y) = e^{xy}\sin(x^2y)$.
$$f(x,y)= {\Sigma_{n=0}^{\infty}} \left[ \left(\begin{matrix} n \\ k \end{matrix}\right)\left. \frac{\partial f}{ \partial x^{n-k}\partial y^k }\right\vert_{M} (x-x_0)^{n-k}(y-y_0)^k \right]$$
I already did all the derivatives, just want to make sure whether my result is correct.
$$f(1;\pi)=\frac{1}{3!}(-4e^{\pi})(x-1)(y-\pi)$$
Your formula for the two variables Taylor expansion has the ${1\over n!}$ factors missing. The correct formula is $$f(x,y)\rightsquigarrow\sum_{n=0}^\infty{1\over n!}\sum_{k=0}^n{n\choose k}{\partial^n f\over\partial x^{n-k}\partial y^k}\biggr|_{(1,\pi)} (x-1)^{n-k}(y-\pi)^k\ .$$ According to this formula the term you want is $${1\over4!}{4\choose 3}{\partial^4 f\over\partial x\partial y^3}\biggr|_{(1,\pi)} (x-1)(y-\pi)^3\ .$$ We therefore have to compute ${\partial^4 f\over\partial x\partial y^3}$. Mathematica obtained $$\eqalign{{\partial^4 f\over\partial x\partial y^3}&=e^{xy} x^2 \bigl((12x-6x^3+5x^2y-7x^4y)\cos(x^2y)\cr &\qquad\qquad+(3-15x^2+xy-9x^3y+2x^5y)\sin(x^2y)\bigr)\ ,\cr}$$ so that $${\partial^4 f\over\partial x\partial y^3}\biggr|_{(1,\pi)}=e^\pi(-6+2\pi)\ .$$ It follows that your term is $$\left({\pi\over3}-1\right)e^\pi(x-1)(y-\pi)^3\ .$$