a) This is a first order linear PDE. We begin by writing equation of characteristic lines:

\begin{equation*}

\frac{dy}{4y}=\frac{dx}{x}

\rightarrow \ln{y}=4\ln{x}+C

\end{equation*}

\begin{equation*}

\Rightarrow \frac{y}{x^4}=C

\end{equation*}

This concludes:

\begin{equation*}

u(x,y)=f(\frac{y}{x^4})

\end{equation*}

where $f: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.

For $u(x,y)$ to be continuous at $(x,y)=(0,0)$, we should have $$u(0,0)=\lim_{x,y\rightarrow0}{u(x,y)}=0$$For this limit to exist, $\lim_{x,y\rightarrow 0}f(\frac{y}{x^4})$ should exist, meaning $f$ should tend to the limit-value regardless of the path on $x$-$y$ plane in which $x$ and $y$ tend to zero. In particular when we choose $y=Cx^4$ for some C, we get $\lim_{x,y\rightarrow 0}{f(\frac{y}{x^4}})=f(C)$. This being true $\forall C\in \mathbb{R}$ concludes $f$ is a constant function. The only continuous function satisfying PDE is the identically constant function.

b) Analogous to part (a), we start with writing equation of characteristic lines:

\begin{equation*}

\frac{dy}{-4y}=\frac{dx}{x}

\rightarrow \ln{y}=-4\ln{x}+C

\end{equation*}

\begin{equation*}\Rightarrow yx^4=C

\end{equation*}

This concludes:

\begin{equation*}

u(x,y)=f(yx^4)\end{equation*}

where $f: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.

This time, $$\lim_{x,y\rightarrow 0}f(yx^4)=\lim_{c \rightarrow 0}f(c)$$

For $u(x,y)$ to be continuous we just need to define $u(0,0)=\lim_{c \rightarrow 0}f(c)$. The easiest way to do is of course by getting $f$ continuous at zero and letting $u(0,0)=f(0)$.