Compute the values of $d y$ and $\Delta y$ for the function $y=e^{3 x}+3 x$ given $x=0$ and $\Delta x=d x=0.05$.

Round your answers to four decimal places, if required. You can use a calculator, spreadsheet, browser, etc. to calculate $d y$ and $\Delta y$.
\[
\begin{array}{l}
d y=\text { Number } \\
\Delta y=\text { Number }
\end{array}
\]

Step 1 ：Given the function \(y = e^{3x} + 3x\), we are asked to compute the values of \(dy\) and \(\Delta y\) at \(x = 0\) with \(\Delta x = dx = 0.05\).

Step 2 ：The differential \(dy\) is given by the derivative of the function times \(dx\), i.e., \(dy = f'(x)dx\).

Step 3 ：The increment \(\Delta y\) is given by the difference in the function's values at \(x + \Delta x\) and \(x\), i.e., \(\Delta y = f(x + \Delta x) - f(x)\).

Step 4 ：First, we need to find the derivative of the function \(f(x) = e^{3x} + 3x\). The derivative is \(f'(x) = 3e^{3x} + 3\).

Step 5 ：Substitute \(x = 0\) and \(dx = 0.05\) into the expression for \(dy\) to find its value. The result is \(dy = 0.3\).

Step 6 ：Substitute \(x = 0\) and \(\Delta x = 0.05\) into the expression for \(\Delta y\) to find its value. The result is \(\Delta y = 0.3118\).

Step 7 ：Final Answer: \[\begin{array}{l}d y=\boxed{0.3} \\Delta y=\boxed{0.3118}\end{array}\]