Find all $\alpha\in\mathbb{R}$ such that $\{(x,y)\in\mathbb{R}^2: \sin(x+y)=\alpha y\}$ is a graph fo some differentiable function Find all $\alpha\in\mathbb{R}$ such that $\{(x,y)\in\mathbb{R}^2: \sin(x+y)=\alpha y\}$ is a graph of some differentiable function $g:\mathbb{R}\rightarrow\mathbb{R}$.

Any hint please? I know the implicit function theorem. In order to use it I'd need to check that 
$$\frac{\partial}{\partial y} (\sin(x+y)-\alpha y) \ne 0$$
But the conclusion of the IFT is only local.  
 A: NOTE: New solution, added later.
Start with the observation that given $\alpha, x,$ there exists $y$ such that $\sin (x+y) = \alpha y.$ Proof: The function $y\to \sin (x+y) -\alpha y$ is continuous and takes arbitrarily large positive and negative values. Apply the IVT.
Case 1: $|\alpha | > 1.$ Claim: The solution set is always the graph of a differentiable function. From the above all we need to show is uniqueness. I.e., for each $x$ there is a unique $y$ that gives a solution. If not, then $\sin(x+y)-\alpha y = 0$ for two distinct values of $y.$ By Rolle, the $y$-derivative, which equals $\cos (x+y)-\alpha$ must be $0$ somewhere. Because $|\alpha | > 1,$ this can't happen.
So for case 1 we've shown the solution set is the graph of a function $g$ on the $x$-axis. We can use the implicit function theorem (which is a local result) to show differentiability (which is a local propery): $\partial (\sin(x+y)-\alpha y)/\partial y = \cos (x+y)-\alpha \ne 0$ everywhere. Thus $g$ is differentiable everywhere by the IFT.
Case 2: $|\alpha | < 1.$ Claim: The solution set is not a graph. Now we know that for any $x$ there are solutions $y.$ The problem is that there are too many of them. If $\alpha = 0$ then for any $x,$ there are infinitely many solutions to $\sin (x +y) = 0.$ Thus the solution set is not a graph.
Suppose $0< \alpha < 1.$ Take $x = 0.$ Define $f(y) = \sin (y) -\alpha y.$ Then $y = 0$ is a solution, but there are others. For small positive $y, \alpha y < \sin y;$ thus $f> 0$ for such $y.$ But $\sin \pi = 0, $ so $f(\pi)<0.$ By the IVT there exists $y> 0$ such that $f(y) = 0.$ Thus for $x= 0,$ there are at least two $y$ values in the solution set, namely $0$ and the positive $y$ we found. So no graph. Same thing for $-1< \alpha < 0.$
Case 3: $|\alpha |=1.$ This is the most interesting case. Claim: The solution set is a graph of some $g(x),$ but $g$ is not differentiable on $\mathbb {R}.$ Take $\alpha = 1.$ Fix $x\in \mathbb {R}.$ Then the function $y \to \sin(x+y) -y$ is continuous and strictly decreasing, and takes arbitrarily large positive and negative values. Hence there is a unique zero $y$ for this function, which implies the solution set for $\alpha =1$ is a graph of some function $g.$
So we have $\sin(x+g(x))=g(x)$ for all $x.$ Note that $g(0)=0.$ Suppose $g'(0)$ exists. Differentiating, we get
$$\cos(0+g(0))(1+g'(0)) = g'(0),$$
a contradiction.
I'll leave the $\alpha = -1$ subcase to the reader.
