Prove that for $(x, y) \in \Bbb R^2$
\begin{cases} x+y+\sin(xy)=2c \\ \sin(x^2 + y) = c^2 \\ \end{cases}
can give a solution for all $c ∈ \Bbb R$ close enough to the origin, by the inverse function theorem.
I am familiar with the IFT, its conditions, and the result. However, I am having a hard time seeing connection between proving the problem and IFT, let alone solving it.
Consider the function $$f(x,y,c):=\bigl(x+y+\sin(xy)-2c, \ \sin(x^2+y^2)-c^2\bigr)\ .$$ We obviously have $f(0,0,0)=(0,0)$. This means that when $c=0$ then the point $(x_0,y_0):=(0,0)$ solves your two equations. The implicit function theorem says that under a certain "technical assumption" we have $$f(x_c,y_c,c)=(0,0)$$ for all $c$ near $0$ and uniquely defined $x_c=\phi(c)$, $\>y_c=\psi(c)$ with $\phi$, $\psi\in C^1$ and $\phi(0)=\psi(0)=0$. Therefore your equations will have a single solution $(x_c,y_c)$ also for $c$ near $0$, whereby $x_c$, $y_c$ depend differentiably on $c$, and $x_0=y_0=0$.
The "technical assumption" is that $f\in C^1$ near $(0,0,0)$ (obviously satisfied), and that $$\det\left[\matrix{{\partial f_1\over\partial x}&{\partial f_1\over\partial y} \cr {\partial f_2\over\partial x}&{\partial f_2\over\partial y}\cr}\right]_{(0,0,0)}\ne0\ .$$ Therefore you have to compute this determinant and to check whether it is $\ne0$.