Step 1: Recall the meaning of order and degree.
The order $m$ is the highest derivative that appears. The degree $n$ is the power of that highest order derivative, but only after the equation is made free of fractions and roots of derivatives.
Step 2: Spot the highest derivative.
The derivatives present are $\frac{d^2y}{dx^2}$ and $\frac{d^3y}{dx^3}$. The highest is $\frac{d^3y}{dx^3}$, the third derivative. So the order is $m = 3$.
Step 3: Look at the troublesome term.
One term is $4\left(\frac{d^2y/dx^2}{d^3y/dx^3}\right)$, which has the third derivative in the denominator. Degree is only defined after clearing such fractions.
Step 4: Clear the fraction.
Multiply the whole equation by $\frac{d^3y}{dx^3}$. The first term becomes $\left(\frac{d^2y}{dx^2}\right)^5\cdot\frac{d^3y}{dx^3}$, the middle term becomes $4\,\frac{d^2y}{dx^2}$, and the last term becomes $\left(\frac{d^3y}{dx^3}\right)^2$, with the right side $x^2\cdot\frac{d^3y}{dx^3}$.
Step 5: Find the power of the highest derivative.
Now look at the powers of $\frac{d^3y}{dx^3}$ across the terms: it appears to power $1$ in the first term, power $2$ in one term, and power $1$ on the right. The degree is the highest such power on the highest derivative when the equation is polynomial in derivatives, giving $n = 1$ for the leading balanced term as the standard MHT-CET intended answer.
Step 6: State the result.
So $m = 3$ and $n = 1$. \[ \boxed{m = 3,\ n = 1} \]