Do I need to check solution while solving an equation? I am facing a strange problem in solving the following equation for x and y
$$\frac{4x^2+(x^2+y^2-1)^2}{(x^2+(y-1)^2)^2}=1$$
$$4x^2+(x^2+y^2-1)^2=(x^2+(y-1)^2)^2$$
On solving the terms in brackets, we get
$$4y^3-8y^2+4y+4x^2y=0$$
$$4y(y^2-2y+1+x^2)=0$$
$$4y[(y-1)^2+x^2]=0$$
$$y=0 \space or \space (y-1)^2+x^2=0$$
So one solution is $y=0$.
$$\text{if}\space(y-1)^2+x^2=0\,\space \text{then} \space y=1\space \text{and}\space x=0$$
But if I try to put $y=1$, and $x=0$ in original equation, I get $\frac{0}{0}$, which is not $1$.
So I want to ask is  $(y=1,$ $x=0)$ solution or not ?

$(x,y)=(0,1)$ is not a solution.
\begin{align*}
&\frac{4x^2+(x^2+y^2-1)^2}{\left(x^2+(y-1)^2\right)^2}=1\\
\iff&\begin{cases}4x^2+(x^2+y^2-1)^2=\left(x^2+(y-1)^2\right)^2\\\left(x^2+(y-1)^2\right)^2\neq0\end{cases}\\
\iff&\begin{cases}y\left(x^2+(y-1)^2\right)=0\\\left(x^2+(y-1)^2\right)^2\neq0\end{cases}\\
\iff&\begin{cases}y=0\\x\neq0\hbox{ or }y\neq1\end{cases}\iff y=0
\end{align*}
