The only such function is the identity function on $R$. Setting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find $f(y)=f(f(y))$ and thus $f(0)=f(f(0))=0$. It follows from (ii) that $f\left(x^{4}+y\right)=x^{3} f(x)+f(y)$ for all $x, y \in \mathbf{R}$. Set $y=0$ to obtain $f\left(x^{4}\right)=x^{3} f(x)$ for all $x \in \mathrm{R}$, and so $f\left(x^{4}+y\right)=f\left(x^{4}\right)+f(y)$ for all $x, y \in \mathbf{R}$. The functional equation suggests that $f$ is additive, that is, $f(a+b)=f(a)+f(b)$ for all $a, b \in \boldsymbol{R}$. We now show this. First assume that $a \geq 0$ and $b \in \boldsymbol{R}$. It follows 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 $f(-x)=\frac{f\left(x^{4}\right)}{(-x)^{3}}=-f(x)$, $x \neq 0$. Since $f$ is odd, we have that, for $a<0$ and $b \in R$, $f(a+b)=-f((-a)+(-b))=-(f(-a)+f(-b))=f(a)+f(b)$. Therefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a, b \in \mathbf{R}$. We now show that $\{s \in R \mid f(s)=0\}=\{0\}$. Recall that $f(0)=0$. Assume that there is a nonzero $h \in \mathrm{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(n h)=0$ or $n h \in\{s \in R \mid f(s)=0\}$ for all $n \in \mathrm{N}$. However, this is a contradiction to the given condition (i). It's now easy to check that $f$ is one-to-one. Assume that $f(a)=f(b)$ for some $a, b \in \operatorname{R}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b \in\{s \in R \mid f(s)=0\}=\{0\}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x \in \mathrm{R}$. This completes the proof.