If $x^2$ $=$ $\frac{9y^6}{8}$ $-$ $\frac{5y^2}{2}$ is the solution of some equation, prove that there is a solution $y=f(x)$ on an interval $I$ such that $f$ is one-to-one on $I$ with initial condition $f(-2)=\sqrt{2}$ and identify the inverse of $f$.
My progress: According to initial condition, I think it is feasible to take the square root, attach minus sign, and assign it for $x$. However, how then can we proceed to prove the injectivity? Wondering whether should we prove $f$ is bijective to deduce the inverse of $f$. Any help would be strongly welcomed, since it is urgent!
You can directly use the inverse function theorem for continuous monotonous functions. In $$ x=g(y)=-y\sqrt{\frac{9y^4}8-\frac52}=-yh(y) $$ on the interval $\sqrt[4]{\frac{20}9}=\sqrt2\sqrt[4]{\frac59}\le y$ the function $g$ is a product of two monotonous functions $-y$ and $h(y)$ with constant signs, thus again a monotonous function. In consequence it is invertible on the range $(-\infty,0]$ for that domain.
You can just as well recover a differential equation for this function. Take the derivative of the equation $$ 2x=\left(\frac{27y^5}4-5y\right)y'(x) $$ and consider this as an ODE with the given initial condition.
Further, using the original equation the above ODE could be replaced with a new one $$ xy=\left(3x^2+5y^2\right)y'(x), $$ which is even less singular as an ODE (it will have in general different solutions than the first, just on the equation in question do these coincide). In the 2nd quadrant containing the point $(x,y)=(-2,\sqrt2)$ this ODE is well-defined, continuous and smooth with a constant sign of the derivative. Any solution of the IVP will thus be locally monotonous.