When we have something like $y = 2x$ we understand $y$ to be the value of the function $f$ at each point $x$ where $f(x) = 2x$, to reiterate, $y$ is not a function but merely a label for the output of $f$. In the case of implicit differentiation we may have something in the form $f(x,y) = 0$ where $f$ is the function and $x, y$ are variables over some domain. I have seen in a lot of introductory calculus books that they write, "to differentiate $f(x,y) = 0$ we use the chain rule and treat $y$ as a function of $x$", now I know that $\frac{dy}{dx}$ is not $0$ in these cases as when we change $x$ there is a resulting change in $y$, i.e the input variables of the function influence each other, but I would like to ask how is it valid to call $y$ as a function of $x$ when it clearly isn't? In addition to this, how does the implicit function theorem come into play here?
Why do we call $y$ a function of $x$ in implicit differentiation
implicit-differentiation
Related Solutions
Depending on how you entered it into WolframAlpha, it is most likely partial differentiating, meaning it considers $x$ and $y$ to be independent variables (specifically, $\frac{\partial y}{\partial x} = 0$). This is why WA's result is different from yours. If you want WA to interpret $y$ as a function of $x$, you have to write y(x), not just y when you enter $y$ into the expression.
Your results looks fine as it is.
With a bit of work, you can show that $R(x,y)$ is only defined for $y=x$ and $y=-x$ excluding the point $(0,0)$. Hence your function does not satisfy the conditions of the implicit function theorem, which states that your function should be at least continuously differentable to apply the method of implicit differentiation.
Note, you could extend your problem to complex numbers, in that case the domain of applicability of $R(x,y)$ would be extended. However, even in that case, it would still not be the entire plane because no definition of the square root or the inverse cosine is analytic on the entire complex plane. In particular, if you choose the most "natural" extension to the complex plane of the inverse cosine, it is not defined on the branch points $-1$ and $1$. Those correspond exactly to the cases $y=x$ and $y=-x$. In other words, naïve implicit differentiation does not work in that case either. You would therefore need analytic extensions of the square root and the inverse cosine such that they are valid in the case $y=x$. But, those extensions will not correspond to the definitions you adopted in the real case.
Best Answer
I had a similar question recently and here's how I understand it.
The implicit function theorem for two dimensions states that for a function
$$F:\mathbb{R}^2\rightarrow \mathbb{R}$$ $$(x,y)\mapsto F(x,y)$$
such that $F$ is continuously differentiable near $(x_0,y_0)$. If $F(x_0,y_0)=0$ and $(\partial_2F)(x_0,y_0)\neq0$, then there exists an open neighbourhood $U$ of $x_0$ and a function $$f:U\rightarrow\mathbb{R}$$That is continuously differentiable such that $f(x_0)=y_0$, $F(x,f(x))=0$ for all $x\in U$ and $f'(x_0)=-\frac{(\partial_1F)(x_0,y_0)}{(\partial_2F)(x_0,y_0)}$.
(Note that $\partial_iF$ stands for the partial derivative of F with respect to it's i-th argument.)
This is why we can treat $y$ as a function of $x$ even though it might be impossible to get an equation of the form $y=g(x)$, the graph of the curve clearly fails the vertical line test, etc. It's just that near every point on the curve, there exists some function whose graph will be the same as the graph of $F(x,y)=0$ near that point and $y=f(x)$ so we can use chain rule to differentiate. (Unless $(\partial_2F)(x_0,y_0)=0$ of course, in which case, there will be a vertical tangent so the derivative is undefined.)