Given y^3+3y=x. Prove that y=f(x) is a single valued function in x.

Consider f(y)= y^3+3y-x

Let us assume x is a particular constant.

Then f'(y)= 3y^2+3 which is always greater than zero.

So since f(y) is a cubic in y, it is bound to have atleast one root, but since it is also an increasing function, hence there will be exactly one root.

Therefore for any real x there exists one and only real root of f(y). So f(y)=0 for single y only.

Or y^3+3y=x is a single valued function.