Q.$\sec^{-1}\dfrac{7x^3-5y^3}{7x^3+5y^3}=m$, show that $\dfrac{d^2y}{dx^2}=0$

Given $\sec^{-1}\left(\dfrac{7x^3-5y^3}{7x^3+5y^3}\right)=m$, a constant.

Step 1 — since $m$ is constant, $\sec(m)$ is some fixed constant $k$, and the fraction itself must equal that constant: $\dfrac{7x^3-5y^3}{7x^3+5y^3}=k$.

Step 2 — cross-multiply: $7x^3-5y^3=k(7x^3+5y^3)=7kx^3+5ky^3$.

Step 3 — collect $x^3$ and $y^3$ terms: $7x^3(1-k)=5y^3(1+k)$, so $\dfrac{y^3}{x^3}=\dfrac{7(1-k)}{5(1+k)}$, a constant. …

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