(1) Send $X$ to $\cos$ and $Y$ to $\sin$. The kernel of this homomorphism consists from the polynomials $f\in\mathbb{R}[X,Y]$ with the property $f(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$. Now prove that this implies $f$ divisible by $X^2+Y^2-1$: consider $f$ as a polynomial in $X$ with coefficients in $\mathbb{R}[Y]$ and write $f=(X^2+Y^2-1)g+r$, where $\deg_Xr\le 1$. Then $r=a+bX$, where $a,b\in\mathbb{R}[Y]$. From $f(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$ we get that $r(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$, that is, $a(\sin t)+b(\sin t)\cos t=0$ for any $t\in\mathbb{R}$. This gives us that $a^2(\sin t)=b^2(\sin t)(1-\sin^2t)$ for any $t\in\mathbb{R}$. But $a,b$ are polynomials and the last relation implies that $a^2(Y)=b^2(Y)(1-Y^2)$ and this is enough to deduce $a=b=0$ (why?).
(2) We show that $x$, the residue class of $X$ in $\mathbb{R}[X,Y]/(X^2+Y^2-1)$ is irreducible and does not divide $1+y$ and $1-y$, the residue classes of $1+Y$ and $1-Y$. (Note that in $\mathbb{R}[x,y]$ we have $x^2=(1+y)(1-y)$.)
Edit. If $x=z_1z_2$, then, by using (4) (see below) we get $N(x)=N(z_1)N(z_2)$ $\Leftrightarrow$ $Y^2-1=N(z_1)N(z_2)$. Now we have the following cases:
(i) $\deg N(z_1)=0\Leftrightarrow z_1\in\mathbb R^*$,
(ii) $\deg N(z_1)=2$ $\Leftrightarrow$ $\deg N(z_2)=0$ $\Leftrightarrow$ $z_2\in\mathbb R^*$,
(iii) $\deg N(z_1)=\deg N(z_2)=1$. If $N(z_1)=Y-1$, then $a_1^2(Y)+b_1^2(Y)(Y^2-1)=Y-1\Rightarrow Y-1\mid a_1(Y)\Rightarrow\exists a_2\in\mathbb R[Y]$ such that $a_1=(Y-1)a_2$ and pluggin this in the foregoing equation we get $a_2^2(Y)(Y-1)+b_1^2(Y)(Y+1)=1$. Looking now at the leading coefficients of $a_2$ and $b_1$ we find that one of these is zero (false!) or the sum of their square is zero (false!).
Assume that $x\mid 1-y$. Then $N(x)\mid N(1-y)\Leftrightarrow Y^2-1\mid (Y-1)^2$, a contradiction.
(3) I've proved here all you need for this part.
(4) As one can see from $(1)$ the elements of $\mathbb{R}[x,y]$ can be uniquely written as $a(y)+b(y)x$. Define a "norm" $N:\mathbb{R}[x,y]\to\mathbb{R}[Y]$ by $N(a(y)+b(y)x)=a^2(Y)+b^2(Y)(Y^2-1)$. $N$ is multiplicative and using this we get that the units of $\mathbb{R}[x,y]$ are the non-zero elements of $\mathbb{R}$.
For $\mathbb{C}[x,y]$ we can see, via the isomorphism from part $(3)$ that the invertible elements are the non-zero elements of $\mathbb{C}$, the powers of $x+iy$ and $x-iy$, and products of the non-zero elements of $\mathbb{C}$ with the powers of $x+iy$ and $x-iy$.
Every function in the image of your map is (or extends to) a continuous function $\mathbb R\to\mathbb R$. Yet there are functions $F\to F$ which are not continuous (or, when $F=\mathbb Q$, do not extend to continuous functions).
This means that the map is not surjective.
Alternatively, every function in the image of your maps is identically zero if it has infinitely many zeroes, yet there exist non-zero functions $F\to F$ which have infinitely many zeroes. This again implies non-surjectivity, and if you look at it correctly, also injectivity.
Best Answer
The minimal polynomial of $a+ib$ is a polynomial of minimal degree $P$ such that $P(a+ib)=0$, if $f(a+ib)=0$, we can write $f=UP+V$ where $\deg(V)<\deg(P)$ (Euclidean division). We deduce that $f(a+ib)=U(a+ib)P(a+ib)+V(a+ib)=V(a+ib)=0$, since $P$ is minimal, $V=0$.
The map $e(f)=f(a+ib)$ factors by $\bar e:\mathbb{R}[X]/(P)\rightarrow \mathbb{C}$.
Suppose that $b=0$, write $P=X-a$ it is the minimal polynomial of $a+ib$ and $e$ is not surjective since $\mathbb{R}[X]/(X-a)\simeq\mathbb{R}$.
Suppose $b\neq 0$
The minimal polynomial of $a+ib$ is $(X-a)^2+b^2$ and the kernel of $f$ is $(P)$, and $f$ is surjective since $\dim_{\mathbb{R}}(\mathbb{R}[X]/(P(X))=2=\dim_{\mathbb{R}}\mathbb{C}$.