Given the curve $y=\frac{5x}{x-3}$. Find its asymptotes, if any. Given the curve $y=\frac{5x}{x-3}$. To examine its asymptotes if any.
We are only taking rectilinear asymptotes in our consideration
My solution goes like this:

We know that, a straight line $x=a$, parallel to $y$ axis can be a vertical asymptote of a branch of the curve $y=f(x)$, iff $f(x)\to\infty$ when $x\to a+0$, $x\to a-0$ or $x\to a$. Similarly, a straight line $y=a$, parallel to $x$ axis can be a horizontal asymptote of a branch of the curve $x=\phi(y)$, iff $x\to\infty$ when $y\to a+0$, $y\to a-0$ or $y\to a$. Using these lemmas, we obtain $x=3$ and $y=5$ as the rectilinear asymptotes. This is because, $y=\frac{5x}{x-3}=f(x)$, then as $x\to 3$, $f(x)\to \infty$. Also, since $y=\frac{5x}{x-3}$, thus,$xy-3y=5x$ or $xy-5x=3y$, hence $x=\frac{3y}{y-5}=\phi(y)$. Due to which $\phi(y)\to \infty$ as $y\to 5$.

Now, I tried checking, if there is any oblique asymptote. We know that, $y=mx+c$, is an oblique asymptote of $y=f(x)$, iff $\exists $ a finite $m=\lim_{|x|\to\infty}\frac{y}{x}$ and $c=\lim_{|x|\to\infty} y-mx$. Now, we have the function $y=\frac{5x}{x-3}$ and hence, $\frac{y}{x}=\frac{5}{x-3}$ and hence, $m=\lim_{|x|\to\infty}\frac{y}{x}=\lim_{|x|\to\infty}\frac{5}{x-3}=0$. Hence, $c=\lim_{|x|\to\infty} y-mx=\frac{5x}{x-3}=\frac{5}{1-\frac 3x}=5$. So, the oblique asymptote of the given curve is $y=mx+c=5$, which is same as the horizontal asymptote, parallel to $x-axis$ as shown above. Is the solution correct? If not, where is it going wrong?

You have correctly found the asymptotes of the function.  Observe that we can express the function in the form
$$f(x) = \frac{5x}{x - 3} = \frac{5x - 15 + 12}{x - 3} = \frac{5x - 15}{x - 3} + \frac{12}{x - 3} = \frac{12}{x - 3} + 5$$
which tells us that the graph of $f$ can be obtained by dilating the graph of $y = 1/x$ by a factor of $12$, shifting it to the right by three units, and up by $5$ units. Since the curve $y = 1/x$ has horizontal asymptote $y = 0$ and vertical asymptote $x = 0$, it follows that the function $f$ has horizontal asymptote $y = 5$ and horizontal asymptote $x = 3$.
A function $f$ has horizontal asymptote $y = k$ if
$$\lim_{x \to \infty} f(x) = k$$
or
$$\lim_{x \to -\infty} f(x) = k$$
Observe that
\begin{align*} 
\lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \frac{4x}{x - 3} = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5\\
\lim_{x \to -\infty} f(x) & \lim_{x \to -\infty} \frac{4x}{x - 3} = = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5
\end{align*}
which tells us that the only horizontal asymptote is $y = 5$.
In fact, a rational function can have at most one horizontal asymptote.  This is not true for all functions.  For instance, the function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = \frac{x}{|x| + 1}$ has two horizontal asymptotes since
\begin{align*}
\lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{x + 1} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} = 1\\
\lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{-x + 1} = \lim_{x \to \infty} \frac{1}{-1 + \frac{1}{x}} = -1
\end{align*}
A function $f$ has a vertical asymptote at $x = a$ if at least one of the following four statements is true:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty & \lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty & \lim_{x \to a^-} f(x) & = -\infty
\end{align*}
In this case,
\begin{align*}
\lim_{x \to 3^+} f(x) & = \lim_{x \to 3^+} \frac{5x}{x - 3} = \lim_{x \to 3^+}  \left(\frac{12}{x - 3} + 5\right) = \infty\\
\lim_{x \to 3^-} f(x) & = \lim_{x \to 3^-} \frac{5x}{x - 3} = \lim_{x \to 3^-}  \left(\frac{12}{x - 3} + 5\right) = -\infty
\end{align*}
so $x = 3$ is a vertical asymptote of the graph.
A rational function can have a horizontal asymptote when the degree of the numerator is at most that of the denominator, which is the case here since both the numerator and denominator have degree one.  A rational function can have an oblique asymptote when the degree of the numerator exceeds that of the denominator by one.  A rational function can have a curvilinear asymptote when the degree of the numerator exceeds that of the denominator by more than one.
