By differentiating implicitly with respect to $x$ both sides of the implicit equation
$$
\begin{equation*}
2x^{2}+y^{2}=33,\tag{1}
\end{equation*}
$$
since the derivatives of both sides should be equal we get successively:
$$
\begin{eqnarray*}
&&\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =\frac{d}{dx}\left( 33\right) \\
&\Rightarrow &\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =0 \\
&\Leftrightarrow &\frac{d}{dx}\left( 2x^{2}\right) +\frac{d}{dx}\left(
y^{2}\right) =0 \\
&\Leftrightarrow &4x+2y\frac{dy}{dx}=0,\qquad \frac{d}{dx}\left(
y^{2}\right) =2y\frac{dy}{dx}\text{ by the chain rule} \\
&\Leftrightarrow &\frac{dy}{dx}=-\frac{4x}{2y}=-\frac{2x}{y}\tag{2} \\
&\Rightarrow &\left. \frac{dy}{dx}\right\vert _{x=2,y=5}=-\frac{4}{5}.\tag{3}
\end{eqnarray*}
$$
The equation of the tangent line at $(2,5)$ is
$$
\begin{equation*}
y-5=-\frac{4}{5}(x-2),\tag{4}
\end{equation*}
$$
while the equation of the normal line to the curve $2x^{2}+y^{2}=33$ at $(2,5)$ is
$$
\begin{equation*}
y-5=\frac{5}{4}(x-2)\Leftrightarrow y=\frac{5}{4}x+\frac{5}{2},\tag{5}
\end{equation*}
$$
because the slope $m$ of the tangent line and the slope $m^{\prime }$ of the normal line are related by $mm^{\prime }=-1$.
ADDED. In a more general case, when we have a differentiable implicit
function $F(x,y)=0$, let $y=f(x)$ denote the function such that $F(x,f(x))\equiv 0\quad$ ($f(x)$ does not need to be explicitly known). If we differentiate both sides of $F(x,y)=0$ and apply the chain rule, we get the following total derivative with respect to $x$:
$$\frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{
\partial F}{\partial y}\frac{dy}{dx}\equiv 0.\tag{A}$$
Solving $(\mathrm{A})$ for $\frac{dy}{dx}$, gives us the following formula
$$\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{
\partial F}{\partial y}.\tag{B}$$
An inflection point is where the second derivative is zero and changes sign. Thus, if the graph is of the second derivative then notice that the zeroes with sign change are at 1 and 7 on that graph.
If the graph is of the first derivative then the local min/max will be the inflection points since this is where if one were to look at the derivative of this graph it would be the second derivative that goes to zero at the extreme point as well as the sign change that is the criteria.
Best Answer
Solving for gradient [i.e. "slope"] at $x = 0$ ($x = 0$ is that is the equation of the y-axis) given
$\;y^3+2y = \sin x+\cos x-1+3x^{2}\tag{1}$
Solve for $y$ at $x = 0$: $\quad x = 0 \implies y^3 + 2y = y(y^2 + 2) = 0\implies y = 0, \;x, y \in \mathbb R$
Use implicit differentiation on the left hand side and the right hand side: as a check, $$\frac{dy}{dx} = \frac{6x + \cos x - \sin x}{3y^2 + 2}$$
Evaluating the derivative at $(0, 0)$ will give the slope/gradient $m$ of the curve at the $y\text{-intercept}.\;$ As a check: we should get $m = 1/2.$
And here, in fact, is the graph of your function near the origin, and a graph of your function with the line tangent to the curve at $(0, 0)$
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