A differentiable function $\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right) \mathrm{satisfies} \mathrm{f}\text{'}\left(\mathrm{x}\right)={\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2+5 \mathrm{and} \mathrm{f}\left(0\right)=1.$ Then the equation of tangent at the point where the curve crosses      -y-axis is

f(0) = 1.

So, the point where curve crosses y-axis is (0, 1).

Slope of tangent at (0, 1) is $\mathrm{f}\text{'}\left(0\right)={\left(\mathrm{f}\left(0\right)\right)}^2+5=6.$

Therefore, the equation of tangent at (0, 1) is

$\begin{array}{l} \mathrm{y}-1=6\left(\mathrm{x}-0\right)\\ \mathrm{or} 6\mathrm{x}-\mathrm{y}+1=0\end{array}$