The only such function is the identity function on $\mathbb{R}$.

Setting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find
$$\begin{equation*} f(y)=f(f(y)) \tag{1} \end{equation*}$$
[1 mark.] and thus $f(0)=f(f(0))=0$ [1 mark.]. It follows from (ii) that $f\left(x^{4}+y\right)= x^{3} f(x)+f(y)$ for all $x, y \in \mathbb{R}$. Set $y=0$ to obtain
$$\begin{equation*} f\left(x^{4}\right)=x^{3} f(x) \tag{2} \end{equation*}$$
for all $x \in \mathbb{R}$, and so
$$\begin{equation*} f\left(x^{4}+y\right)=f\left(x^{4}\right)+f(y) \tag{3} \end{equation*}$$
for all $x, y \in \mathbb{R}$. The functional equation (3) suggests that $f$ is additive, that is, $f(a+b)= f(a)+f(b)$ for all $a, b \in \mathbb{R}$. [1 mark.] We now show this.

First assume that $a \geq 0$ and $b \in \mathbb{R}$. It follows from (3) that
$$f(a+b)=f\left(\left(a^{1 / 4}\right)^{4}+b\right)=f\left(\left(a^{1 / 4}\right)^{4}\right)+f(b)=f(a)+f(b)$$
We next note that $f$ is an odd function, since from (2)
$$f(-x)=\frac{f\left(x^{4}\right)}{(-x)^{3}}=\frac{f\left(x^{4}\right)}{-x^{3}}=-f(x), \quad x \neq 0$$
Since $f$ is odd, we have that, for $a<0$ and $b \in \mathbb{R}$,
$$\begin{aligned} f(a+b) & =-f((-a)+(-b))=-(f(-a)+f(-b)) \\ & =-(-f(a)-f(b))=f(a)+f(b) \end{aligned}$$
Therefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a, b \in \mathbb{R}$. [2 marks.]

We now show that $\{s \in \mathbb{R} \mid f(s)=0\}=\{0\}$. Recall that $f(0)=0$. Assume that there is a nonzero $h \in \mathbb{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(n h)=0$ or $n h \in\{s \in \mathbb{R} \mid f(s)=0\}$ for all $n \in \mathbb{N}$. However, this is a contradiction to the given condition (i). [1 mark.]

It's now easy to check that $f$ is one-to-one. Assume that $f(a)=f(b)$ for some $a, b \in \mathbb{R}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b \in\{s \in \mathbb{R} \mid f(s)=0\}=\{0\}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x \in \mathbb{R}$. [1 mark.] This completes the proof.

Again, the only such function is the identity function on $\mathbb{R}$.

As in Solution 1, we first show that $f(f(y))=f(y)$, $f(0)=0$, and $f\left(x^{4}\right)=x^{3} f(x)$. [2 marks.] From the latter follows
$$f(x)=0 \Longrightarrow f\left(x^{4}\right)=0$$
and from condition (i) we get that $f(x)=0$ only possibly for $x \in\{0,1,-1\}$. [1 mark.]

Next we prove
$$f(a)=b \Longrightarrow f(\sqrt[4]{|a-b|})=0$$
This is clear if $a=b$. If $a>b$ then
$$\begin{aligned} f(a) & =f((a-b)+b)=(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(f(b)) \\ & =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(b) \\ & =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(f(a)) \\ & =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(a) \end{aligned}$$
so $(a-b)^{3 / 4} f(\sqrt[4]{a-b})=0$ which means $f(\sqrt[4]{|a-b|})=0$. If $a<b$ we get similarly
$$\begin{aligned} f(b) & =f((b-a)+a)=(b-a)^{3 / 4} f(\sqrt[4]{b-a})+f(f(a)) \\ & =(b-a)^{3 / 4} f(\sqrt[4]{b-a})+f(b) \end{aligned}$$
and again $f(\sqrt[4]{|a-b|})=0$. [2 marks.]

Thus $f(a)=b \Longrightarrow|a-b| \in\{0,1\}$. Suppose that $f(x)=x+b$ for some $x$, where $|b|=1$.
Then from $f\left(x^{4}\right)=x^{3} f(x)$ and $f\left(x^{4}\right)=x^{4}+a$ for some $|a| \leq 1$ we get $x^{3}=a / b$, so $|x| \leq 1$.
Thus $f(x)=x$ for all $x$ except possibly $x= \pm 1$. [1 mark.] But for example,
$$f(1)=f\left(2^{4}-15\right)=2^{3} f(2)+f(f(-15))=2^{3} \cdot 2-15=1$$
and
$$f(-1)=f\left(2^{4}-17\right)=2^{3} f(2)+f(f(-17))=2^{3} \cdot 2-17=-1$$
[1 mark.] This finishes the proof.