The implicit function theorem says that, under some regularity hypotheses, there is a function $f(x)$ such that the solution set to $F(x,y)=0$ near a particular solution $(x_0,y_0)$ exists and is given by $y=f(x)$. It requires you to have such a point $(x_0,y_0)$ (in your problem they are telling you that it is $(0,0)$) and to check that $\frac{\partial F}{\partial y}(x_0,y_0) \
eq 0$.
In contrast to some of the other answers, you do not need a full Taylor development to prove the statement as given in the OP. All you need is to check that $\left. \frac{\partial}{\partial y} \left ( e^{\sin(xy)} + x^2 -2y - 1 \right ) \right |_{(x,y)=(0,0)} \
eq 0$. (Alternately, for just existence of solution you could find $x=f(y)$, which would let you take $\frac{\partial}{\partial x}$ instead of $\frac{\partial}{\partial y}$.)