Answer
One $x$ value corresponds to two $y$ value, so the function doesn't exist. It is not a function.
Work Step by Step
To determine what we are asked, we first have to try to simplify and solve the equation.
By first look we can notice that it looks like a quadratic equation.
$2xy-5y^2=4$
Let's simplify to make it look more clear (put the terms to the right-hand side):
$5y^2-2xy+4=0$
$ax^2+bx+c=0$
Comparing our equation to the general form of equation, we see:
$a=5$
$b=-2x$
$c=4$
Let's just solve the quadratic equation:
$D=b^2-4ac=4x^2-80$
$y=\frac{-b±\sqrt{D}}{2a}=\frac{2x±\sqrt{4x^2-80}}{10}=\frac{x±\sqrt{x^2-20}}{5}$
So we have two solutions
$y_1=\frac{x-\sqrt{x^2-20}}{5}$
$y_2=\frac{x+\sqrt{x^2-20}}{5}$
One $x$ value corresponds to two $y$ value, so the function doesn't exist. It is not a function.
