Answer
$y=-\dfrac{37}{4}$
Work Step by Step
Squaring both sides of the given equation, $
\sqrt{y+3}-\sqrt{y-3}=1
,$ results to
\begin{array}{l}\require{cancel}
(\sqrt{y+3}-\sqrt{y-3})^2=(1)^2
\\
(\sqrt{y+3})^2-2(\sqrt{y+3})(\sqrt{y-3})+(\sqrt{y-3})^2=1
\\
y+3-2\sqrt{y+3}\sqrt{y-3}+y-3=1
\\
(y+y)+(3-3-1)=2\sqrt{y+3}\sqrt{y-3}
\\
2y+1=2\sqrt{y+3}\sqrt{y-3}
.\end{array}
Squaring both sides for the second time results to
\begin{array}{l}\require{cancel}
(2y+1)^2=(2\sqrt{y+3}\sqrt{y-3})^2
\\
(2y)^2+2(2y)(1)+(1)^2=4(y+3)(y-3)
\\
4y^2+4y+1=4(y^2-9)
\\
4y^2+4y+1=4y^2-36
\\
(4y^2-4y^2)+4y=-36-1
\\
4y=-37
\\
y=-\dfrac{37}{4}
.\end{array}
Upon checking, $
y=-\dfrac{37}{4}
,$ satisfies the original equation.