so the formula you want to apply reads \[ F(a+h, b+k) = \sum_{i,j=0}^{n} \frac 1{i!j!}\cdot \frac{\partial^{i+j}F}{\partial x^i\partial y^j}(a,b)h^ik^j + \sum_{i+j = n+1} \frac 1{i!j!} \cdot \frac{\partial^2F}{\partial x^i\partial y^j}(a+\theta h, b + \theta k)h^ik^j \] and tells you how $F$ behaves in a neighbourhood of $(a,b)$. The corresponding theorem says the for each $(h,k)$ there is some $\theta \in [0,1]$ such that this formula holds.
Here we have \begin{align*} F(x,y) &= \sin x \sin y\\\ F(0,0) &= 0\\\ \frac{\partial F}{\partial x}(x,y) &= \cos x \sin y\\\ \frac{\partial F}{\partial x}(0,0) &= 0\\\ \frac{\partial F}{\partial y}(x,y) &= \sin x \cos y\\\ \frac{\partial F}{\partial y}(0,0) &= 0\\\ \frac{\partial^2 F}{\partial x^2}(x,y) &= -\sin x \sin y\\\ \frac{\partial^2 F}{\partial x^2}(0,0) &= 0\\\ \frac{\partial^2 F}{\partial x\partial y}(x,y) &= \cos x \cos y\\\ \frac{\partial^2 F}{\partial x\partial y}(0,0) &= 1\\\ \frac{\partial^2 F}{\partial y^2}(x,y) &= -\sin x \sin y\\\ \frac{\partial^2 F}{\partial y^2}(0,0) &= 0\\\ \end{align*} So in the first sum above all but one Term are zero, the non-vanishing term is \[ \frac 1{1!1!}\cdot \frac{\partial^2 F}{\partial x\partial y}(0,0)hk = hk\] Now the third partial derivatives \begin{align*} \frac{\partial^3 F}{\partial x^3}(x,y) &= -\cos x \sin y\\\ \frac{\partial^3 F}{\partial x^2\partial y}(x,y) &= -\sin x \cos y\\\ \frac{\partial^3 F}{\partial x\partial y^2}(x,y) &= -\cos x \sin y\\\ \frac{\partial^3 F}{\partial y^3}(x,y) &= -\sin x \cos y\\\ \end{align*} Pluging in, we get \begin{align*} \sum_{i+j = n+1} &\frac 1{i!j!} \cdot \frac{\partial^2F}{\partial x^i\partial y^j}(a+\theta h, b + \theta k)h^ik^j \\ &= -\frac 16\cdot\cos\theta h\sin\theta k\cdot h^3 - \frac 12\cdot \sin\theta h \cos \theta k \cdot h^2k\\\ &\quad{}- \frac 12\cos\theta h \sin \theta k\cdot hk^2 - \frac 16 \cdot \sin\theta h \cos \theta k \cdot k^3\\\ &= -\frac 16 h(h^2 + 3k^2)\cos\theta h\sin \theta k -\frac 16 k(3h^2 +k^2) \sin\theta h\cos\theta k. \end{align*} .. as we wanted.