Show that for $(a,b,c)\in \mathbb{R}^3$, the system of equations has a unique solution for $(a,b,c)$ sufficiently close to $(0,0,0)$ 
Let $\phi(u,v): \mathbb{R}^2 \to \mathbb{R}$ be a smooth function such that $\phi(0,0)=0$ and $\partial_u \phi(0,0) = \partial_v\phi(0,0) = 0. $ Show that for $(a,b,c)\in \mathbb{R}^3$, the system of equations $$\sin(x) +\phi(y,z) = a$$$$\sin(y)+\phi(x,z) = b$$$$\sin(z)+\phi(x,y) = c$$ has a unique solution for $(a,b,c)$ sufficiently close to $(0,0,0)$.

I have no idea about how to solve this problem, I am thinking about by using the inverse function theorem, and calculating a determinant of the Jacobian matrix. But I am not sure if it's correct.
Could you please help me? Thank you!
 A: 
Implicit function theorem:
Let $U \subset \mathbb{R}^{n+m}$ and let $F$:$U \rightarrow \mathbb{R}^n, F \in C^1$ be a function for which $F(c) = 0$ and the derivative of $F$ at point $c$ is surjective. Then there exists an environment of $c$ where the equation $F(c)=0$ implicitly defines $n$ passive variables as a function of $m$ active variables. Such function $g$ is unique and continuously differentiable.

Let $F:\mathbb{R}^{3+3} \rightarrow \mathbb{R}^3$ be a function so that $$F(a,b,c,x,y,z) = (\sin(x) + \phi(y,z) - a, \sin(y) +\phi(x,z)-b, \sin(z) +\phi(x,y)-c)$$ and let $c=(0,0,0,0,0,0)$. Now $F(c) = (0,0,0)$.
Writing the derivative in matrix form we get
$$
DF(c) =  \left( \begin{array}{cccccc}
-1 & 0 & 0 & \cos0 & \frac{\partial{\phi}}{\partial u}(0,0) & \frac{\partial{\phi}}{\partial v}(0,0) \\
0 & -1 & 0 & \frac{\partial{\phi}}{\partial u}(0,0) & \cos0 & \frac{\partial{\phi}}{\partial v}(0,0) \\
0 & 0 & -1 & \frac{\partial{\phi}}{\partial u}(0,0) & \frac{\partial{\phi}}{\partial v}(0,0) & \cos0  \end{array} \right) 
=
\left( \begin{array}{cccccc}
-1 & 0 & 0 & 1 & 0 & 0 \\
0 & -1 & 0 & 0 & 1 & 0 \\
0 & 0 & -1 & 0 & 0 & 1 \end{array} \right) 
$$
We can see that the derivative is surjective. By the implicit function theorem there exists an environment of $c$ where we can express $a$, $b$ and $c$ as a function of $x$, $y$ and $z$: $(a,b,c) = g(x,y,z)$. In such environment we have $F(g(x,y,z),x,y,z) = 0$ and so $(a,b,c) = g(x,y,z)$ is the unique solution to the system of equations. 
