The solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x+y)^2$ is

A normal is drawn at a point $\mathrm{P}(x, y)$ of a curve $y=\mathrm{f}(x)$. The normal meets the $X$ axis at $Q$. $l(\mathrm{PQ})=\mathrm{k} \cdot(\mathrm{k}$ is a constant) Then equation of the curve through $(0, k)$ is

If $\mathrm{f}(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}, x \neq 0$ is continuous at $x=0$, then the value of $\mathrm{f}(0)$ is