Let $P(x,y)$ denote the proposition $f(x^2yf(x))+f(1)=x^2f(x)+f(y).$ Note that if $f$ has a fixed point $a$ then $P(a,1)$ gives $f(a^3)=a^3,$ so $a^3$ is a fixed point as well. $P(x,1)$ implies that $x^2f(x)$ is a fixed point for every $x.$ From $P(1,1)$ we see that $f(1)$ is a fixed point and so is $f(1)^3.$ From $P(1,f(1))$ and $P(1,f(1)^2)$ we get $f(1)^3=f(f(1)^3)=f(f(1)^2)=f(f(1))=f(1)$ so $f(1)=\pm 1.$ -- If $f(1)=1$ then $P(a,y)$ gives $Q(x,y): \ \ f(a^3y)=a^3-1+f(y)$ for all $y\ne 0$ and all fixed points $a.$ We have $f(a^6)=2a^3-1$ and $f(a^9)=a^3-1+f(a^6)=3a^3-2.$ Moreover $a^3$ is a fixed point and so is $a^9.$ So, $a^9=3a^3-2$ and so $a=1, \ \sqrt[3]{-2}.$ But if $f(\sqrt[3]{-2})=\sqrt[3]{-2}$ then $f(-2)=-2$ which cannot be a fixed point. So the only possible fixed point of $f$ is $1.$ Thus $x^2f(x)=1, $ for all $x$ that is $f(x)=\frac{1}{x^2}$ for all $x$ which is a solution. -- If $f(1)=-1$ then $P(a,y)$ gives $Q(x,y): \ \ f(a^3y)=a^3+1+f(y)$ for all $y\ne 0$ and all fixed points $a.$ We have $f(a^6)=2a^3+1$ and $f(a^9)=a^3-1+f(a^6)=3a^3+2.$ Moreover $a^3$ is a fixed point and so is $a^9.$ So, $a^9=3a^3+2$ and so $a=-1, \ \sqrt[3]{2}.$ But if $f(\sqrt[3]{2})=\sqrt[3]{2}$ then $f(2)=2$ which cannot be a fixed point. So the only possible fixed point of $f$ is $-1.$ Thus $x^2f(x)=-1, $ for all $x$ that is $f(x)=-\frac{1}{x^2}$ for all $x$ which is a solution.

I like this one. 1. $P(x,1)$ gives $x^2f(x)$ is a fixed point, so $u$ fixed point implies $u^3$ is a fixed point. Also, $f(1)$ is a fixed point. 2. Use $P(1,y)$, so $f(f(1)y)=f(y)$, so $f(1)^3=f(f(1)^3)=f(f(1)^2)=f(f(1))=f(1)$ giving $f(1)=\pm 1$. Assume $f(1)=1$. For any fixed point $u$, $P(u,y)$ gives $f(u^3y)=u^3+f(y)-1$. Spam $y=u^3$ and $u^6$. Also note that $u^9$ is also a fixed point. Then we have $u^9=3u^3-2$, so $u^3=1, -2$. Therefore, only possible $u$ is $1$. Therefore, $x^2f(x)=1$, or $f(x) = \frac{1}{x^2}$. Similarly, if $f(1)=-1$, we have $f(x)=-\frac{1}{x^2}$. Done.

#4 solved a cubic incorrectly. Let $c=\sqrt[3]2$, and $P(x,y)$ will denote that assertion $f(x^2yf(x))+f(1)=x^2f(x)+f(y)$. $P(1,x)\Rightarrow f(xf(1))=f(x)$ $P(f(1),1)-P(1,1)\Rightarrow f(1)=\pm1$ $\textbf{Case 1: }f(1)=1$ $P(x,1)\Rightarrow f(x^2f(x))=x^2f(x)$ If $f(k)=k$, then: $P(k,1)\Rightarrow f(k^3)=k^3\Rightarrow f(k^9)=k^9$ $P(k,k^3)\Rightarrow f(k^6)=2k^3-1$ $P(k,k^6)\Rightarrow f(k^9)=3k^3-2\Rightarrow k^9-3k^3+2=0\Rightarrow k\in\{-c,1\}\Rightarrow f(x)\in\left\{\frac1{x^2},-\frac c{x^2}\right\}$. Assume that for some nonzero $a$ we have $f(a)=-\frac c{a^2}$. $P(a,-c)\Rightarrow f(c^2)=1-2c$ If $f(c^2)=\frac1{c^4}$, then we get a contradiction, and if $f(c^2)=-\frac c{c^4}$, we also get a contradiction. So $\boxed{f(x)=\frac1{x^2}}$ holds for all $x\ne0$, and this works. $\textbf{Case 2: }f(1)=-1$ Similar proof to Case 1 with the solution being $\boxed{f(x)=-\frac1{x^2}}$, omitted.

$$f(x^2yf(x))+f(1)=x^2f(x)+f(y)$$$P(x,1) \Rightarrow f(x^2f(x))=x^2f(x))$ $x \Rightarrow x^2f(x) $ $f(x^6f(x)^3)=x^6f(x)^3$ $P(x,x^4f(x)^2) \Rightarrow f(x^6f(x)^3)+f(1)=x^2f(x)+f(x^4f(x)^2)$ $P(x,x^2f(x)) \Rightarrow f(x^4f(x)^2)+f(1)=2x^2f(x)$ $$f(1)=\pm1$$$$x^6f(x)^3=3x^2f(x)-2f(1)$$$f(1)=1$ $$(x^2f(x)-1)^2(x^2f(x)+2)=0$$if exist $a$ $f(a)=\frac{-2}{a^2}$ $P(a,y)$ $f(-2y)=f(y)-3$ $f(-2)=-2$ condtradiction,reasons $f(-2)\in (\frac{1}{4},-\frac{1}{2})$ answer:$f(x)=\pm\frac{1}{x^2}$